Sequence & Series [409 marks]

1. 23M.1.AHL.TZ1.10
Consider the arithmetic sequence 𝑢1, 𝑢2, 𝑢3, … .
The sum of the first 𝑛 terms of this sequence is given by 𝑆𝑛 = 𝑛 2 + 4𝑛.
[N/A]
[[N/A]]
(a.i) Find the sum of the first five terms.
[2]
Markscheme

recognition that 𝑛 = 5 (M1) 𝑆5 = 45 A1 [2 marks]

(a.ii) Given that 𝑆6 = 60, find 𝑢6.

[2]
Markscheme

METHOD 1 recognition that 𝑆5 + 𝑢6 = 𝑆6 (M1) 𝑢6 = 15 A1 METHOD 2

6
recognition that 60 = 2 (𝑆1 + 𝑢6 ) (M1) 60 = 3(5 + 𝑢6 ) 𝑢6 = 15 A1 METHOD 3
substituting their 𝑢1 and 𝑑 values into 𝑢1 + (𝑛 − 1)𝑑 (M1)
𝑢6 = 15 A1

[2 marks]

(b) Find 𝑢1.

[2]
Markscheme

recognition that 𝑢1 = 𝑆1 (may be seen in (a)) OR substituting their 𝑢6 into 𝑆6 (M1) OR

6
equations for 𝑆5 and 𝑆6 in terms of 𝑢1 and 𝑑 1 + 4 OR 60 = 2 (𝑈1 + 15) 𝑢1 = 5 A1 [2
marks]

(c) Hence or otherwise, write an expression for 𝑢𝑛 in terms of 𝑛.

[3]
Markscheme

EITHER valid attempt to find 𝑑 (may be seen in (a) or (b)) (M1) 𝑑 = 2 (A1) OR
valid attempt to find 𝑆𝑛 − 𝑆𝑛−1 (M1) 𝑛 2 + 4𝑛 − (𝑛 2 − 2𝑛 + 1 + 4𝑛 − 4) (A1) OR
𝑛
equating 𝑛 2 + 4𝑛 = 2 (5 + 𝑢𝑛 ) (M1) 2𝑛 + 8 = 5 + 𝑢𝑛 (or equivalent) (A1) THEN
𝑢𝑛 = 5 + 2(𝑛 − 1) OR 𝑢𝑛 = 2𝑛 + 3 A1

[3 marks]

Consider a geometric sequence, 𝑣𝑛 , where 𝑣2 = 𝑢1 and 𝑣4 = 𝑢6.

(d) Find the possible values of the common ratio, 𝑟.
[3]
Markscheme
 recognition that 𝑣2 𝑟 2 = 𝑣4 OR (𝑣3 )2 = 𝑣2 × 𝑣4 (M1) 𝑟 2 = 3 OR 𝑣3 = (±)5√3
(A1) 𝑟 = ±√3 A1 Note: If no working shown, award M1A1A0 for √3. [3 marks]

(e) Given that 𝑣99 < 0, find 𝑣5.

[2]
Markscheme
45
recognition that 𝑟 is negative (M1) 𝑣5 = −15√3 (= − ) A1 [2 marks]
√3

2. 23M.3.AHL.TZ2.2
This question asks you to examine linear and quadratic functions constructed in systematic ways using
arithmetic sequences.
Consider the function 𝐿(𝑥) = 𝑚𝑥 + 𝑐 for 𝑥 ∈ ℝ where 𝑚, 𝑐 ∈ ℝ and 𝑚, 𝑐 ≠ 0.
Let 𝑟 ∈ ℝ be the root of 𝐿(𝑥) = 0.
If 𝑚, 𝑟 and 𝑐, in that order, are in arithmetic sequence then 𝐿(𝑥) is said to be an AS-linear function.
[N/A]
[[N/A]]
(a) Show that 𝐿(𝑥) = 2𝑥 − 1 is an AS-linear function.
[2]
Markscheme
1 1 1 1 3
𝑚 = 2, 𝑐 = −1 𝑟 = 2 (A1) 2, 2, −1 EITHER 𝑑 (= 2 − 2 = −1 − 2) = − 2 A1 OR
3
this sequence has a common difference of − 2 A1 OR the (arithmetic) mean of 2 and −1 is
1
A1 THEN hence 𝐿(𝑥) = 2𝑥 − 1 is an AS-linear function AG [2 marks]
2

Consider 𝐿(𝑥) = 𝑚𝑥 + 𝑐.
𝑐
(b.i) Show that 𝑟 = − 𝑚.
[1]
Markscheme
𝑐
(𝐿(𝑟) = 0 ⇒)𝑚𝑟 + 𝑐 = 0 A1 𝑟 = − 𝑚 AG Note: Award A0 for numerical
verification from 𝐿(𝑥) = 2𝑥 − 1 in part (a). [1 mark]

𝑚2
(b.ii) Given that 𝐿(𝑥) is an AS-linear function, show that 𝐿(𝑥) = 𝑚𝑥 − 𝑚+2.
[4]
Markscheme

METHOD 1 EITHER attempts to use (𝑑 =)𝑟 − 𝑚 = 𝑐 − 𝑟 (M1) Note: Award (M1) for
𝑐 𝑐
attempting to use (𝑑 =)𝑚 − 𝑟 = 𝑟 − 𝑐. (𝑑 =) − 𝑚 − 𝑚 = 𝑐 − (− 𝑚) A1 Note: Award
𝑐 𝑐−𝑚
A1 for (𝑑 =) − 𝑚 − 𝑚 = . removes the denominator 𝑚 from their expression involving 𝑚
2
2 𝑚+𝑐
and 𝑐 (M1) 𝑚 + 𝑐𝑚 + 2𝑐 = 0 (or equivalent) OR attempts to use =𝑟 (M1)
2
2𝑐
𝑚+𝑐 =−𝑚 A1 removes the denominator 𝑚 from their expression involving 𝑚 and
 𝑐 (M1) 𝑚2 + 𝑐𝑚 + 2𝑐 = 0 (or equivalent) OR attempts to use 𝑐 = 𝑚 + 2𝑑 (M1)
𝑐 𝑐
𝑐 = 𝑚 + 2 (− 𝑚 − 𝑚) A1 Note: Award A1 for 𝑐 = 𝑚 + 2 (𝑐 − (− 𝑚)). removes the
denominator 𝑚 from their expression involving 𝑚 and 𝑐 (M1) 𝑚2 + 𝑐𝑚 + 2𝑐 = 0 (or
equivalent) OR attempts to use 𝑟 = 𝑚 + 𝑑 and 𝑐 = 𝑚 + 2𝑑 (𝑐 = 𝑚 + 2(𝑟 − 𝑚)) (M1)
2 𝑐−𝑚
𝑚 + 𝑑𝑚 + 𝑚 + 2𝑑 = 0 (or equivalent) A1 substitutes 𝑑 = 2 into their expression
involving 𝑚 and 𝑑 (M1) 𝑚 + 𝑐𝑚 + 2𝑐 = 0 (or equivalent) THEN 𝑐(𝑚 + 2) = −𝑚2 ⇒
2
𝑚2 𝑚2
𝑐 = − 𝑚+2 A1 Note: Award A1 for a convincing demonstration that 𝑐 = − 𝑚+2. so
𝑚2
𝐿(𝑥) = 𝑚𝑥 − 𝑚+2 AG Note: Do not accept working backwards from the AG. METHOD
2 considers 𝐿(𝑥) = 𝑚𝑥 − 𝑚𝑟 attempts to use (𝑑 =)𝑟 − 𝑚 = 𝑐 − 𝑟 (M1) Note: Award
(M1) for attempting to use (𝑑 =)𝑚 − 𝑟 = 𝑟 − 𝑐. (𝑑 =)𝑟 − 𝑚 = −𝑚𝑟 − 𝑟 A1 attempts to
𝑚 𝑚2
express 𝑟 in terms of 𝑚 (M1) 2𝑟 + 𝑚𝑟 = 𝑚 ⇒ 𝑟 = 𝑚+2 A1 so 𝐿(𝑥) = 𝑚𝑥 − 𝑚+2
AG Note: Do not accept working backwards from the AG. [4 marks]

(b.iii) State any further restrictions on the value of 𝑚.

[1]
Markscheme

𝑚 ≠ −2 (𝑚 ≠ 0) A1 [1 mark]

There are only three integer sets of values of 𝑚, 𝑟 and 𝑐, that form an AS-linear function. One of these is
𝐿(𝑥) = −𝑥 − 1.
(c) Use part (b) to determine the other two AS-linear functions with integer values of 𝑚, 𝑟 and 𝑐.
[3]
Markscheme

attempts to find an integer value of 𝑚 (M1) e.g. uses the result that 𝑚 + 2 exactly divides 2
OR uses a table OR uses a graph and slider OR uses systematic trial and error Note: Award
𝑚2
(M1) for solving 𝑚2 = 𝑘(𝑚 + 2) for 𝑚 or solving 𝑚𝑟 − 𝑚+2 = 0 for 𝑚 or solving 𝑚2 + 𝑐𝑚 +
2𝑐 = 0 for 𝑚. 𝑚 = −4 OR 𝑚 = −3 (A1) −4, 2, 8 OR −3, 3, 9 𝐿(𝑥) = −4𝑥 +
8, 𝐿(𝑥) = −3𝑥 + 9 A1 Note: Award (M1)(A1)A0 for −4𝑥 + 8 and −3𝑥 + 9. [3 marks]

Consider the function 𝑄(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 for 𝑥 ∈ ℝ where 𝑎 ∈ ℝ, 𝑎 ≠ 0 and 𝑏, 𝑐 ∈ ℝ.

Let 𝑟1, 𝑟2 ∈ ℝ be the roots of 𝑄(𝑥) = 0.
(d) Write down an expression for
[[N/A]]
(d.i) the sum of roots, 𝑟1 + 𝑟2, in terms of 𝑎 and 𝑏.
[1]
Markscheme
𝑏
− A1 [1 mark]
𝑎

(d.ii) the product of roots, 𝑟1 𝑟2, in terms of 𝑎 and 𝑐.

[1]
Markscheme
 𝑐
A1 [1 mark]
𝑎

If 𝑎, 𝑟1, 𝑏, 𝑟2 and 𝑐, in that order, are in arithmetic sequence, then 𝑄(𝑥) is said to be an AS-quadratic function.

(e) Given that 𝑄(𝑥) is an AS-quadratic function,

[[N/A]]
(e.i) write down an expression for 𝑟2 − 𝑟1 in terms of 𝑎 and 𝑏;
[1]
Markscheme

𝑏−𝑎 A1 Note: Award marks as appropriate in parts (e) (ii) and (iii) for use of 𝑟1 − 𝑟𝑏 =
𝑎 − 𝑏. [1 mark]

𝑎2 −𝑎𝑏−𝑏
(e.ii) use your answers to parts (d)(i) and (e)(i) to show that 𝑟1 = .
2𝑎
[2]
Markscheme

𝑏 𝑎2 −𝑎𝑏−𝑏
attempts to eliminate 𝑟2 M1 2𝑟1 = − 𝑎 − (𝑏 − 𝑎) ⇒ 2𝑟1 = (or equivalent)
𝑎
𝑎2 −𝑎𝑏−𝑏
A1 Note: Award A1 for a correct alternative form of ±𝑟1 or ±2𝑟1. so 𝑟1 = AG
2𝑎
Note: Do not accept working backwards from the AG. [2 marks]

1
(e.iii) use the result from part (e)(ii) to show that 𝑏 = 0 or 𝑎 = − 2.
[3]
Markscheme
𝑎+𝑏
METHOD 1 EITHER (𝑟1 =) (A1) attempts to equate two expressions for either 𝑟1 or
2
𝑎+𝑏 𝑎2 −𝑎𝑏−𝑏 𝑎2 −𝑎𝑏−𝑏
2𝑟1 in terms of 𝑎 and 𝑏 M1 = OR 𝑎 + 𝑏 = OR 𝑏 − 𝑟1 = 𝑟1 −
2 2𝑎 𝑎
𝑎2 −𝑎𝑏−𝑏 𝑎2 −𝑎𝑏−𝑏
𝑎 (A1) attempts to use 𝑏 − 𝑟1 = 𝑟1 − 𝑎 with 𝑟1 = M1 𝑏 − ( )=
2𝑎 2𝑎
𝑎2 −𝑎𝑏−𝑏 𝑎2 −𝑎𝑏−𝑏
2𝑎
− 𝑎 OR (𝑟1 =)𝑎 + 𝑑 (A1) attempts to use 𝑟1 = 𝑎 + 𝑑 with 𝑟1 = 2𝑎
and 𝑑 =
𝑏−𝑎 𝑎2 −𝑎𝑏−𝑏 𝑏−𝑎
M1 2𝑎 = 𝑎 + 2 THEN 2𝑎2 + 2𝑎𝑏 = 2𝑎2 − 2𝑎𝑏 − 2𝑏 OR 𝑎 + 𝑎𝑏 = 𝑎2 − 2
2
𝑎𝑏 − 𝑏 4𝑎𝑏 + 2𝑏 = 0 OR 2𝑎𝑏 + 𝑏 = 0 2𝑏(2𝑎 + 1) = 0 OR 𝑏(2𝑎 + 1) = 0 A1 Note:
Award (A1)M1 for any valid approach that correctly leads to 2𝑎𝑏 + 𝑏 = 0 (or equivalent).
1
Do not accept numerical verification from the AG. so 𝑏 = 0 or 𝑎 = − 2 AG METHOD
2 (𝑏 =)𝑎 + 2𝑑 OR (𝑟1 =)𝑎 + 𝑑 (A1) attempts to equate two expressions for either 𝑟1 or
𝑎2 −𝑎(𝑎+2𝑑)−(𝑎+2𝑑)
2𝑟1 in terms of 𝑎 and 𝑑 M1 𝑎 + 𝑑 = OR 2(𝑎 + 𝑑) =
2𝑎
𝑎2 −𝑎(𝑎+2𝑑)−(𝑎+2𝑑)
2𝑎2 + 4𝑎𝑑 + 𝑎 + 2𝑑 = 0 (2𝑎 + 1)(𝑎 + 2𝑑) = 0 A1 Note: Do not
𝑎
1
accept numerical verification from the AG. so 𝑏 = 0 or 𝑎 = − 2 AG [3 marks]

Consider the case where 𝑏 = 0.

(f) Determine the two AS-quadratic functions that satisfy this condition.
[5]
 Markscheme
𝑎 𝑎 𝑎
METHOD 1 𝑟1 = 2 OR 𝑟2 = − 2 OR 𝑑 = − 2 (A1) 𝑐 = −𝑎 (A1) attempts to find
𝑎
the values of 𝑎 (M1) EITHER the roots of 𝑎𝑥 2 − 𝑎 = 0 are ±1 and 2 = ±1 OR
𝑎 𝑎3 𝑐 𝑎2 𝑎3
substitutes 𝑥 = ± into 𝑎𝑥 2 − 𝑎 = 0 giving − 𝑎 = 0 OR (𝑟1 𝑟2 =) = − ⇒𝑐= and
2 4 𝑎 4 4
𝑎3 𝑎3
so −𝑎 = − ⇒ − 𝑎 = 0 Note: Award (M1) for attempting to find the values of 𝑎 from their
4 4
𝑎3 𝑎 𝑎 𝑎3
arithmetic sequence expressed in terms of 𝑎. OR 𝑐 − 𝑟2 = 𝑟2 − 𝑏 ⇒ − 4 − (− 2) = − 2 ⇒ 4 −
𝑎 = 0 THEN 𝑎 = ±2 (A1) (𝑟1 = ±1, 𝑏 = 0, 𝑟2 = ∓1, 𝑐 = ∓2) so 𝑄(𝑥) = 2𝑥 2 −
2, 𝑄(𝑥) = −2𝑥 2 + 2 A1 Note: Award A0 for 2𝑥 2 − 2 and −2𝑥 2 + 2.
Award (A1)(A1)(M1)(A0)A0 for obtaining either 𝑎 = 2 or 𝑎 = −2. METHOD 2 𝑟1 = −𝑑 OR
𝑟2 = 𝑑 OR 𝑎 = −2𝑑 (A1) 𝑐 = 2𝑑 (A1) attempts to find the values of 𝑑 (M1)
EITHER the roots of −2𝑑𝑥 2 + 2𝑑 = 0 are ±1 OR substitutes 𝑥 = ±𝑑 into −2𝑑𝑥 2 + 2𝑑 = 0
𝑐 2𝑑
giving −2𝑑 3 + 2𝑑 = 0 OR attempts to use 𝑟1 𝑟2 = 𝑎 to form −𝑑 2 = −2𝑑 Note: Award (M1)
for attempting to find the values of 𝑑 from their arithmetic sequence expressed in terms of 𝑑.
THEN 𝑑 = ±1 (A1) (𝑎 = ±2, 𝑟1 = ±1, 𝑏 = 0, 𝑟2 = ∓1, 𝑐 = ∓2) so 𝑄(𝑥) = 2𝑥 2 −
2
2, 𝑄(𝑥) = −2𝑥 + 2 A1 Note: Award A0 for 2𝑥 2 − 2 and −2𝑥 2 + 2.
Award (A1)(A1)(M1)(A0)A0 for obtaining either 𝑑 = 1 or 𝑑 = −1. METHOD 3 𝑎 = 2𝑟1 OR
𝑟2 = −𝑟1 OR 𝑑 = −𝑟1 (A1) 𝑐 = −2𝑟1 (A1) attempts to find the values of 𝑟1
(M1) EITHER the roots of 2𝑟1 𝑥 − 2𝑟1 = 0 are ±1 OR substitutes 𝑥 = ±𝑟1 into 2𝑟1 𝑥 2 − 2𝑟1 =
2
𝑐 −2𝑟
0 giving 2𝑟1 3 − 2𝑟1 = 0 OR attempts to use 𝑟1 𝑟2 = 𝑎 to form −𝑟1 2 = 2𝑟 1 Note: Award (M1)
1
for attempting to find the values of 𝑟1 from their arithmetic sequence expressed in terms of 𝑟1.
THEN 𝑟1 = ±1 (A1) (𝑎 = ±2, 𝑟1 = ±1, 𝑏 = 0, 𝑟2 = ∓1, 𝑐 = ∓2) so 𝑄(𝑥) = 2𝑥 2 −
2, 𝑄(𝑥) = −2𝑥 2 + 2 A1 Note: Award A0 for 2𝑥 2 − 2 and −2𝑥 2 + 2.
Award (A1)(A1)(M1)(A0)A0 for obtaining either 𝑟1 = 1 or 𝑟1 = −1. [5 marks]

1
Now consider the case where 𝑎 = − 2.
(g.i) Find an expression for 𝑟1 in terms of 𝑏.
[2]
Markscheme
1
attempts to express 𝑟1 in terms of 𝑏 with 𝑎 = − 2. (M1) Note: Do not award (M1) if 𝑎 =
1 𝑎+𝑏 𝑎2 −𝑎𝑏−𝑏
is used. EITHER uses 𝑟1 = OR uses 𝑟1 = OR uses 𝑟1 − 𝑎 = 𝑏 − 𝑟1 THEN
2 2 2𝑎
1
2𝑏−1 𝑏 1 𝑏−
2
𝑟1 = 4
(= 2 − 4 , = 2
) A1 [2 marks]

(g.ii) Hence or otherwise, determine the exact values of 𝑏 and 𝑐 such that AS-quadratic functions are formed.
−𝑝±𝑞√𝑠
Give your answers in the form where 𝑝, 𝑞, 𝑠 ∈ ℤ+.
2
[5]
Markscheme
1
METHOD 1 EITHER substitutes their expression for 𝑟1 with 𝑎 = − 2 into 𝑄(𝑥)(= 0)
2𝑏−1 1 2𝑏−1 2 2𝑏−1 6𝑏+1 3𝑏 1
(M1) 𝑄 ( ) (= 0) ⇒ − 2 ( ) +𝑏( ) + 𝑐(= 0) OR 𝑟2 = (= + 4)
4 4 4 4 2
 1 6𝑏+1
substitutes their expression for 𝑟2 with 𝑎 = − 2 into 𝑄(𝑥)(= 0) (M1) 𝑄 ( ) (= 0) ⇒
4
1 6𝑏+1 2 6𝑏+1 4𝑏+1 1
−2( ) +𝑏( ) + 𝑐(= 0) THEN 𝑐 = (= 2𝑏 + 2) (seen anywhere) A1
4 4 2
−5±2√5
4𝑏2 + 20𝑏 + 5 = 0 attempts to solve their quadratic in 𝑏 (M1) 𝑏 = A1
2
4𝑏+1 −9±4√5
substitutes into 𝑐 = 𝑐= A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as
2 2
decimal values. Note: Award a maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i).
1
METHOD 2 substitutes their expressions for 𝑟1 and 𝑟2 with 𝑎 = − 2 into 𝑄(𝑥) (M1)
1 2𝑏−1 6𝑏+1 1 3 1 1 4𝑏+1 1
− 2 (𝑥 − ( )) (𝑥 − ( )) − 2 𝑥 2 + 𝑏𝑥 − 8 𝑏2 + 8 𝑏 + 32 𝑐 = (= 2𝑏 + 2) (seen
4 4 1
1 3 1 1
anywhere) A1 2𝑏 + = − 𝑏2 + 𝑏 + 4𝑏2 + 20𝑏 + 5 = 0 attempts to solve their
2 8 8 32
−5±2√5 4𝑏+1 −9±4√5
quadratic in 𝑏 (M1) 𝑏 = 2 A1 substitutes into 𝑐 = 2 𝑐 = 2 A1
Note: Award A0A0 for 𝑏 and 𝑐 expressed as decimal values. Note: Award a maximum of
6𝑏+1 3𝑏 1
(M1)A1(M1)A0A0FT for FT from part (g) (i). METHOD 3 𝑟2 = 4 (= 2 + 4) substitutes
1 𝑐 2𝑏−1 6𝑏+1 𝑐
their expressions for 𝑟1 and 𝑟2 with 𝑎 = − into 𝑟1 𝑟2 = (M1) ( )( )= 1 (or
2 𝑎 4 4 −2
4𝑏+1 1
equivalent) EITHER 𝑐 = (= 2𝑏 + 2) (seen anywhere) A1
2
1 2𝑏−1 6𝑏+1 6𝑏+1 2𝑏−1 1
OR − 2 ( )( )−( )= − (− 2) A1
4 4 4 4
−5±2√5
THEN 4𝑏2 + 20𝑏 + 5 = 0 attempts to solve their quadratic in 𝑏 (M1) 𝑏 = 2
4𝑏+1 −9±4√5
A1 substitutes into 𝑐 = 2 𝑐 = 2 A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as
decimal values. Note: Award a maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i).
1 −𝑏±√𝑏2 +2𝑐
METHOD 4 attempts to equate two expressions for 𝑟1 with 𝑎 = − 2 (M1) =
−1
2𝑏−1 2𝑏+1 4𝑏+1 1
(±√𝑏2 + 2𝑐 = )𝑐= (= 2𝑏 + 2) (seen anywhere) A1 12𝑏2 − 4𝑏 − 1 +
4 4 2
1
32 (2𝑏 + 2) = 0 (4𝑏2 + 20𝑏 + 5 = 0) attempts to solve their quadratic in 𝑏 (M1) 𝑏 =
−5±2√5 4𝑏+1 −9±4√5
A1 substitutes into 𝑐 = 2 𝑐 = 2 A1 Note: Award A0A0 for 𝑏 and 𝑐
2
expressed as decimal values. Note: Award a maximum of (M1)A1(M1)A0A0FT for FT from
1
part (g) (i). METHOD 5 EITHER 𝑟1 = 𝑑 − 2 substitutes their expression for 𝑟1 in terms of
1 1 1 1 2 1
𝑑with 𝑎 = − 2 into 𝑄(𝑥)(= 0) (M1) 𝑄 (𝑑 − 2) (= 0) ⇒ − 2 (𝑑 − 2) + 𝑏 (𝑑 − 2) +
1 1
𝑐(= 0) OR 𝑟2 = 3𝑑 − 2 substitutes their expression for 𝑟2 in terms of 𝑑 with 𝑎 = − 2 into
1 1 1 2 1
𝑄(𝑥)(= 0) (M1) 𝑄 (3𝑑 − 2) (= 0) ⇒ − 2 (3𝑑 − 2) + 𝑏 (3𝑑 − 2) + 𝑐(= 0) THEN 𝑏 =
1 1
2𝑑 − 2 and 𝑐 = 4𝑑 − 2 (seen anywhere) A1 4𝑑 2 + 8𝑑 − 1 = 0 attempts to solve their
−2±√5 −5±2√5 4𝑏+1
quadratic in 𝑑 (M1) 𝑑 = 𝑏= A1 substitutes into 𝑐 = 𝑐=
2 2 2
−9±4√5
A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as decimal values. Note: Award a
2
1
maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i). METHOD 6 𝑟1 = 𝑑 − 2 and 𝑟2 =
1 1 𝑐
3𝑑 − substitutes their expression for 𝑟1 and 𝑟2 in terms of 𝑑 with 𝑎 = − into 𝑟1 𝑟2 =
2 2 𝑎
1 1 𝑐 1
(M1) (𝑑 − 2) (3𝑑 − 2) = 1 (or equivalent) 𝑐 = 4𝑑 − 2 4𝑑 2 + 8𝑑 − 1 = 0 attempts to solve
−2
−2±√5 −5±2√5 4𝑏+1
their quadratic in 𝑑 (M1) 𝑑 = 𝑏= A1 substitutes into 𝑐 = 𝑐=
2 2 2
 −9±4√5
A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as decimal values. Note: Award a
2
maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i). [5 marks]

3. 22N.3.AHL.TZ0.1
In this question you will investigate series of the form
𝑛
𝛴 𝑖 𝑞 = 1𝑞 + 2𝑞 + 3𝑞 + ⋯ + 𝑛 𝑞 where 𝑛, 𝑞 ∈ ℤ+
𝑖=1
and use various methods to find polynomials, in terms of 𝐧, for such series.
When 𝑞 = 1, the above series is arithmetic.
𝑛 1
(a) Show that 𝛴 𝑖 = 2 𝑛(𝑛 + 1).
𝑖=1
[1]
Markscheme

EITHER
𝑛
𝑆𝑛 = (2 × 1 + (𝑛 − 1) × 1)
2
A1

OR
𝑢1 = 1 and either 𝑢𝑛 = 𝑛 or 𝑑 = 1 stated explicitly A1

OR
1 + 2 + ⋯ + 𝑛 (or equivalent) stated explicitly A1

THEN
𝑛
𝑆𝑛 = 2 (1 + 𝑛) AG

Note: Award A0 for a numerical verification.

[1 mark]

Consider the case when 𝑞 = 2.

𝑛
The following table gives values of 𝑛 2 and 𝛴 𝑖 2 for 𝑛 = 1, 2, 3.
𝑖=1

(b.i) Write down the value of 𝑝.

[1]
Markscheme

14 A1

[1 mark]
(b.ii) The sum of the first 𝑛 square numbers can be expressed as a cubic polynomial with three terms:
𝑛
𝛴 𝑖 2 = 𝑎1 𝑛 + 𝑎2 𝑛2 + 𝑎3 𝑛 3 where 𝑎1 , 𝑎2 , 𝑎3 ∈ ℚ+ .
𝑖=1
Hence, write down a system of three linear equations in 𝑎1 , 𝑎2 and 𝑎3.
[3]
Markscheme

𝑎1 + 𝑎2 + 𝑎3 = 1 A1
2𝑎1 + 4𝑎2 + 8𝑎3 = 5 A1
3𝑎1 + 9𝑎2 + 27𝑎3 = 14 A1

Note: For the third A mark, award A1FT for 3𝑎1 + 9𝑎2 + 27𝑎3 = 𝑝 where 𝑝 is their answer to
part (b) (i).

[3 marks]

(b.iii) Hence, find the values of 𝑎1 , 𝑎2 and 𝑎3.

[2]
Markscheme
1 1
𝑎1 = 6 (= 0.166666 … ≈ 0.167), 𝑎2 = 2 (= 0.5),
1
𝑎3 = 3 (= 0.333333 … ≈ 0.333) A2

Note: Award A1 if only two of 𝑎1 , 𝑎2 , 𝑎3 are correct.

Only award FT if three linear equations, each in 𝑎1 , 𝑎2 and 𝑎3 are stated in part (b) (ii) or (iii).
Award A2FT for their 𝑎1 , 𝑎2 and 𝑎3.
Award A1FT for their 𝑎1 , 𝑎2 and 𝑎3 = 0.

[2 marks]

You will now consider a method that can be generalized for all values of 𝑞.
Consider the function 𝑓(𝑥) = 1 + 𝑥 + 𝑥 2 + ⋯ + 𝑥 𝑛 , 𝑛 ∈ ℤ+.
(c) Show that 𝑥𝑓′(𝑥) = 𝑥 + 2𝑥 2 + 3𝑥 3 + ⋯ + 𝑛𝑥 𝑛 .
[1]
Markscheme

𝑓′(𝑥 ) = 1 + 2𝑥 + 3𝑥 2 + ⋯ + 𝑛𝑥 𝑛−1 A1
𝑛
Note: Award A1 for 𝑓′(𝑥) = 𝛴 𝑖𝑥 𝑖−1.
𝑖=1

⇒ 𝑥𝑓′(𝑥) = 𝑥 + 2𝑥 2 + 3𝑥 3 + ⋯ + 𝑛𝑥 𝑛 AG

[1 mark]

Let 𝑓1 (𝑥) = 𝑥𝑓′(𝑥) and consider the following family of functions:

𝑓2 (𝑥) = 𝑥𝑓1 ′(𝑥)
𝑓3 (𝑥) = 𝑥𝑓2 ′(𝑥)
𝑓4 (𝑥 ) = 𝑥𝑓3 ′(𝑥 )
…
𝑓𝑞 (𝑥) = 𝑥𝑓𝑞−1 ′(𝑥)
𝑛
(d.i) Show that 𝑓2 (𝑥) = 𝛴 𝑖 2 𝑥 𝑖 .
𝑖=1
[2]
Markscheme

METHOD 1
𝑓2 (𝑥) = 𝑥𝑓1 ′(𝑥)
𝑓1 ′(𝑥) = 12 + 22 𝑥 + (32 𝑥 2 ) + ⋯ + 𝑛 2 𝑥 𝑛−1 (= 1 + 4𝑥 + (9𝑥 2 ) + ⋯ + 𝑛2 𝑥 𝑛−1 )
A1

Note: Award A1 for 𝑥𝑓1 ′(𝑥) = 𝑥(12 + 22 𝑥 + (32 𝑥 2 ) + ⋯ + 𝑛 2 𝑥 𝑛−1 ) (= 𝑥(1 + 4𝑥 + (9𝑥 2 ) +
⋯ + 𝑛 2 𝑥 𝑛−1 ))

𝑥𝑓1 ′(𝑥) = 12 𝑥 + 22 𝑥 2 + (32 𝑥 3 ) + ⋯ + 𝑛 2 𝑥 𝑛 (= 𝑥 + 4𝑥 2 + (9𝑥 3 ) + ⋯ + 𝑛 2 𝑥 𝑛 )

A1
𝑛 𝑛
Note: Award A1 for 𝑓1 ′(𝑥) = 𝛴 𝑖 2𝑥 𝑖−1 and A1 for 𝑥𝑓1 ′(𝑥) = 𝑥 𝛴 𝑖 2𝑥 𝑖−1.
𝑖=1 𝑖=1
The second A1 is dependent on the first A1.
Award a maximum of A0A1 if a general term is not considered.
𝑛
= 𝛴 𝑖 2 𝑥𝑖 AG
𝑖=1

METHOD 2
𝑑
𝑓2 (𝑥) = 𝑥 (𝑥𝑓′(𝑥))
𝑑𝑥
= 𝑥(𝑓′(𝑥) + 𝑥𝑓″(𝑥)) (= 𝑥𝑓′(𝑥) + 𝑥 2 𝑓″(𝑥)) A1
𝑛 𝑛
= 𝑥 𝛴 𝑖𝑥 𝑖−1 + 𝑥 2 𝛴 𝑖(𝑖 − 1)𝑥 𝑖−2
𝑖=1 𝑖=1
𝑛 𝑛 𝑛
= 𝛴 𝑖𝑥 𝑖 + 𝛴 𝑖(𝑖 − 1)𝑥 𝑖 (= 𝛴 (𝑖 + 𝑖 2 − 𝑖)𝑥 𝑖 ) A1
𝑖=1 𝑖=1 𝑖=1
𝑛
= 𝛴 𝑖 2 𝑥𝑖 AG
𝑖=1

[2 marks]

𝑛
(d.ii) Prove by mathematical induction that 𝑓𝑞 (𝑥) = 𝛴 𝑖 𝑞 𝑥 𝑖 , 𝑞 ∈ ℤ+ .
𝑖=1
[6]
Markscheme

consider 𝑞 = 1
𝑛
𝑓1 (𝑥) = 𝑥 + 2𝑥 2 + ⋯ 𝑛𝑥 𝑛 (reference to part (c)) and 𝑓1 (𝑥) = 𝛴 𝑖𝑥 𝑖 R1
𝑖=1
𝑛
assume true for 𝑞 = 𝑘, (𝑓𝑘 (𝑥) = 𝛴 𝑖 𝑘 𝑥 𝑖 ) M1
𝑖=1

Note: Do not award M1 for statements such as “let 𝑞 = 𝑘” or “𝑞 = 𝑘 is true”. Subsequent marks
after this M1 are independent of this mark and can be awarded.
 consider 𝑞 = 𝑘 + 1
𝑓𝑘+1 (𝑥) = 𝑥𝑓𝑘 ′(𝑥) M1
𝑛
= 𝑥 𝛴 𝑖 𝑘+1 𝑥 𝑖−1 OR 𝑥(1 + 2𝑘+1 𝑥 + 3𝑘+1 𝑥 2 + ⋯ + 𝑛 𝑘+1 𝑥 𝑛−1 ) A1
𝑖=1

𝑛 𝑛
Note: Award the above M1 if 𝑓𝑘+1 (𝑥) = 𝑥 𝛴 𝑖 𝑘+1 𝑥 𝑖−1 or 𝑥𝑓𝑘 ′(𝑥) = 𝑥 𝛴 𝑖 𝑘+1 𝑥 𝑖−1 (or
𝑖=1 𝑖=1
equivalent) is stated.
𝑛
= 𝛴 𝑖 𝑘+1 𝑥 𝑖 OR 𝑥 + 2𝑘+1 𝑥 2 + 3𝑘+1 𝑥 3 + ⋯ + 𝑛 𝑘+1 𝑥 𝑛 A1
𝑖=1
since true for 𝑞 = 1 and true for 𝑞 = 𝑘 + 1 if true for 𝑞 = 𝑘, hence true for all 𝑞(∈ ℤ+ )
R1

Note: To obtain the final R1, three of the previous five marks must have been awarded.

[6 marks]

(d.iii) Using sigma notation, write down an expression for 𝑓𝑞 (1).

[1]
Markscheme

𝑓𝑞 (1) = 1𝑞 + 2𝑞 + 3𝑞 + ⋯ + 𝑛 𝑞
𝑛 𝑛
= 𝛴 𝑖 𝑞 (= 𝛴 1𝑖 𝑖 𝑞 ) A1
𝑖=1 𝑖=1

[1 mark]

𝑥 𝑛+1 −1
(e) By considering 𝑓(𝑥) = 1 + 𝑥 + 𝑥 2 + ⋯ + 𝑥 𝑛 as a geometric series, for 𝑥 ≠ 1, show that 𝑓(𝑥) = .
𝑥−1
[2]
Markscheme
𝑢 (𝑟𝑛 −1)
uses 𝑆𝑛 = 1 𝑟−1 where 𝑟 = 𝑥 and 𝑢1 = 1 M1
clear indication there are (𝑛 + 1) terms R1
𝑥 𝑛+1 −1
𝑓(𝑥) = AG
𝑥−1

[2 marks]

𝑛𝑥 𝑛+2 −(𝑛+1)𝑥 𝑛+1+𝑥

(f) For 𝑥 ≠ 1, show that 𝑓1 (𝑥) = (𝑥−1)2
.
[3]
Markscheme

METHOD 1
𝑓1 (𝑥) = 𝑥𝑓′(𝑥)
(𝑥−1)(𝑛+1)𝑥 𝑛 −1×(𝑥 𝑛+1 −1)
=𝑥 (𝑥−1)2
M1A1
 Note: Award M1 for attempting to use the quotient or the product rule to find 𝑓′(𝑥).

(𝑛𝑥+𝑥−𝑛−1)𝑥 𝑛 −(𝑥 𝑛+1 −1) 𝑛𝑥 𝑛+1 −𝑛𝑥 𝑛 −𝑥 𝑛 +1

=𝑥 (𝑥−1)2
(= 𝑥 (𝑥−1)2
) A1
Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

𝑛𝑥 𝑛+2 −(𝑛+1)𝑥 𝑛+1 +𝑥

= (𝑥−1)2
AG

METHOD 2
attempts to form (𝑥 − 1)𝑓1 (𝑥) M1
(𝑥 − 1)𝑓1 (𝑥) = 𝑛𝑥 𝑛+1 − (𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛 )
1 𝑥 𝑛+1 −1
𝑓1 (𝑥) = 𝑥−1 (𝑛𝑥 𝑛+1 − ( − 1)) A1
𝑥−1

1 𝑛𝑥 𝑛+1 (𝑥−1)−𝑥 𝑛+1 +𝑥 1 𝑛𝑥 𝑛+2 −𝑛𝑥 𝑛+1 −𝑥 𝑛+1 +𝑥

𝑓1 (𝑥) = 𝑥−1 ( ) (= 𝑥−1 ( )) A1
𝑥−1 𝑥−1

Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

𝑛𝑥 𝑛+2 −(𝑛+1)𝑥 𝑛+1 +𝑥

𝑓1 (𝑥) = (𝑥−1)2
AG

[3 marks]

(g.i) Show that lim 𝑓1 (𝑥) is in indeterminate form.

𝑥→1
[1]
Markscheme
𝑛−(𝑛+1)+1 0
lim 𝑓1 (𝑥) = (= 0) R1
𝑥→1 0

Note: Only award R1 for sufficient simplification of the numerator, for example, as shown
above.
Do not award R1 if lim is not referred to or stated.
𝑥→1

[1 mark]

1
(g.ii) Hence, by applying l’Hôpital’s rule, show that lim 𝑓1 (𝑥) = 𝑛(𝑛 + 1).
𝑥→1 2
[5]
Markscheme

attempts to differentiate both the numerator and the denominator M1

𝑛(𝑛+2)𝑥 𝑛+1 −(𝑛+1)2 𝑥 𝑛 +1
lim 2(𝑥−1)
A1
𝑥→1

𝑛(𝑛+2)𝑥 𝑛+1 −𝑛(𝑛+1)𝑥 𝑛 −(𝑛+1)𝑥 𝑛+1

Note: Award A1 for ((lim ) ). This form can be used in
𝑥→1 2(𝑥−1)

subsequent work.
 (l’Hôpital’s rule applies again since)
𝑛(𝑛+2)𝑥 𝑛+1 −(𝑛+1)2 𝑥 𝑛 +1 0
lim =0 R1
𝑥→1 2(𝑥−1)

Note: Do not award R1 if lim is not referred to or stated. Subsequent marks are independent of
𝑥→1
this R mark.

attempts to differentiate both the numerator and the denominator M1

𝑛(𝑛 + 2)(𝑛 + 1)𝑥 𝑛 − 𝑛(𝑛 + 1)2 𝑥 𝑛−1
lim
𝑥→1 2
𝑛(𝑛 + 2)(𝑛 + 1) − 𝑛(𝑛 + 1)2 𝑛 3 + 3𝑛 2 + 2𝑛 − (𝑛 3 + 2𝑛2 + 𝑛)
= (= )
2 2
𝑛(𝑛+1)((𝑛+2)−(𝑛+1)) 𝑛2 +𝑛
= (= ) A1
2 2
1
= 2 𝑛(𝑛 + 1) AG

[5 marks]

4. 22M.1.AHL.TZ1.10
1
Consider the series ln 𝑥 + 𝑝 ln 𝑥 + 3 ln 𝑥 + ⋯, where 𝑥 ∈ ℝ, 𝑥 > 1 and 𝑝 ∈ ℝ, 𝑝 ≠ 0.
Consider the case where the series is geometric.
1
(a.i) Show that 𝑝 = ± .
√3
[2]
Markscheme

EITHER
attempt to use a ratio from consecutive terms M1
1
𝑝 ln 𝑥 ln 𝑥 1 1
3
= 𝑝 ln 𝑥 OR ln 𝑥 = (ln 𝑥)𝑟 2 OR 𝑝 ln 𝑥 = ln 𝑥 (3𝑝)
ln 𝑥 3

1
Note: Candidates may use ln 𝑥 1 + ln 𝑥 𝑝 + ln 𝑥 3 + ⋯ and consider the powers of 𝑥 in geometric
sequence
1
𝑝 3
Award M1 for 1 = 𝑝 .

OR
1
𝑟 = 𝑝 and 𝑟 2 = 3 M1

THEN
1 1
𝑝2 = 3 OR 𝑟 = ± A1
√3
1
𝑝=± AG
√3

1 1
Note: Award M0A0 for 𝑟 2 = 3 or 𝑝2 = 3 with no other working seen.

[2 marks]
(a.ii) Hence or otherwise, show that the series is convergent.
[1]
Markscheme

EITHER
1 1
since, |𝑝| = and <1 R1
√3 √3

OR
1
since, |𝑝| = and −1 < 𝑝 < 1 R1
√3

THEN
⇒ the geometric series converges. AG

Note: Accept 𝑟 instead of 𝑝.

Award R0 if both values of 𝑝 not considered.

[1 mark]

(a.iii) Given that 𝑝 > 0 and 𝑆∞ = 3 + √3, find the value of 𝑥.

[3]
Markscheme
ln 𝑥
1 (= 3 + √3) (A1)
1−
√3
3 √3
ln 𝑥 = 3 − + √3 − OR ln 𝑥 = 3 − √3 + √3 − 1 (⇒ ln 𝑥 = 2) A1
√3 √3
𝑥 = e2 A1

[3 marks]

Now consider the case where the series is arithmetic with common difference 𝑑.
2
(b.i) Show that 𝑝 = 3.
[3]
Markscheme

METHOD 1
attempt to find a difference from consecutive terms or from 𝑢2 M1
correct equation A1
1 1
𝑝 ln 𝑥 − ln 𝑥 = 3 ln 𝑥 − 𝑝 ln 𝑥 OR 3 ln 𝑥 = ln 𝑥 + 2(𝑝 ln 𝑥 − ln 𝑥)

1
Note: Candidates may use ln 𝑥 1 + ln 𝑥 𝑝 + ln 𝑥 3 + ⋯ and consider the powers of 𝑥 in arithmetic
sequence.
1
Award M1A1 for 𝑝 − 1 = − 𝑝
3

4 4
2𝑝 ln 𝑥 = 3 ln 𝑥 (⇒ 2𝑝 = 3) A1
2
𝑝=3 AG
 METHOD 2
𝑢 +𝑢
attempt to use arithmetic mean 𝑢2 = 1 2 3 M1
1
ln 𝑥+3ln 𝑥
𝑝 ln 𝑥 = A1
2
4 4
2𝑝 ln 𝑥 = 3 ln 𝑥 (⇒ 2𝑝 = 3) A1
2
𝑝=3 AG

METHOD 3
attempt to find difference using 𝑢3 M1
1 1
ln 𝑥 = ln 𝑥 + 2𝑑 (⇒ 𝑑 = − ln 𝑥)
3 3
1 1 1
𝑢2 = ln 𝑥 + 2 (3 ln 𝑥 − ln 𝑥) OR 𝑝 ln 𝑥 − ln 𝑥 = − 3 ln 𝑥 A1
2
𝑝 ln 𝑥 = 3 ln 𝑥 A1
2
𝑝=3 AG

[3 marks]

(b.ii) Write down 𝑑 in the form 𝑘 ln 𝑥, where 𝑘 ∈ ℚ.

[1]
Markscheme
1
𝑑 = − 3 ln 𝑥 A1

[1 mark]

1
(b.iii) The sum of the first 𝑛 terms of the series is ln (𝑥 3).
Find the value of 𝑛.
[8]
Markscheme

METHOD 1
𝑛 1
𝑆𝑛 = ⌊2 ln 𝑥 + ⌊𝑛 − 1⌋ × ⌊− ln 𝑥⌋⌋
2 3
1
attempt to substitute into 𝑆𝑛 and equate to ln (𝑥 3) (M1)
𝑛 1 1
⌊2 ln 𝑥 + ⌊𝑛 − 1⌋ × ⌊− ln 𝑥⌋⌋ = ln ( 3 )
2 3 𝑥
1
ln (𝑥 3) = −ln 𝑥 3 (= ln 𝑥 −3 ) (A1)
= −3 ln 𝑥 (A1)
correct working with 𝑆𝑛 (seen anywhere) (A1)
𝑛 𝑛 1 𝑛(𝑛−1) 𝑛 4−𝑛
⌊2 ln 𝑥 − 3 ln 𝑥 + 3 ln 𝑥⌋ OR 𝑛 ln 𝑥 − 6 ln 𝑥 OR (ln 𝑥 + ( ) ln 𝑥)
2 2 3
correct equation without ln 𝑥 A1
𝑛 7 𝑛 𝑛(𝑛−1)
( − 3 ) = −3 OR 𝑛 − 6 = −3 or equivalent
2 3
 1 𝑛 7 𝑛
Note: Award as above if the series 1 + 𝑝 + 3 + ⋯ is considered leading to 2 (3 − 3 ) = −3.

attempt to form a quadratic = 0 (M1)

𝑛 2 − 7𝑛 − 18 = 0
attempt to solve their quadratic (M1)
(𝑛 − 9)(𝑛 + 2) = 0
𝑛=9 A1

METHOD 2
1
ln (𝑥 3) = −ln 𝑥 3 (= ln 𝑥 −3 ) (A1)
= −3 ln 𝑥 (A1)
listing the first 7 terms of the sequence (A1)
2 1 1 2
ln 𝑥 + ln 𝑥 + ln 𝑥 + 0 − ln 𝑥 − ln 𝑥 − ln 𝑥 + ⋯
3 3 3 3
recognizing first 7 terms sum to 0 M1
4
8 term is − 3 ln 𝑥
th
(A1)
5
9th term is − 3 ln 𝑥 (A1)
sum of 8 and 9 term = −3 ln 𝑥
th th (A1)
𝑛=9 A1

[8 marks]

5. 22M.3.AHL.TZ1.1
This question asks you to explore some properties of polygonal numbers and to determine and prove
interesting results involving these numbers.

A polygonal number is an integer which can be represented as a series of dots arranged in the shape of a regular
polygon. Triangular numbers, square numbers and pentagonal numbers are examples of polygonal numbers.
For example, a triangular number is a number that can be arranged in the shape of an equilateral triangle. The
first five triangular numbers are 1, 3, 6, 10 and 15.
The following table illustrates the first five triangular, square and pentagonal numbers respectively. In each case
the first polygonal number is one represented by a single dot.
For an 𝑟-sided regular polygon, where 𝑟 ∈ ℤ+ , 𝑟 ≥ 3, the 𝑛th polygonal number 𝑃𝑟 (𝑛) is given by
(𝑟−2)𝑛2 −(𝑟−4)𝑛
𝑃𝑟 (𝑛) = , where 𝑛 ∈ ℤ+ .
2
(4−2)𝑛2−(4−4)𝑛
Hence, for square numbers, 𝑃4 (𝑛) = = 𝑛2.
2
𝑛(𝑛+1)
(a.i) For triangular numbers, verify that 𝑃3 (𝑛) = .
2
[2]
Markscheme

(3−2)𝑛 2−(3−4)𝑛 𝑛 2 −(−𝑛)

𝑃3 (𝑛) = OR 𝑃3 (𝑛) = A1
2 2
𝑛2 +𝑛
𝑃3 (𝑛) = A1
2

𝑛2 +𝑛
Note: Award A0A1 if 𝑃3 (𝑛) = 2 only is seen.
Do not award any marks for numerical verification.

𝑛(𝑛+1)
so for triangular numbers, 𝑃3 (𝑛) = AG
2

[2 marks]

(a.ii) The number 351 is a triangular number. Determine which one it is.
[2]
Markscheme

METHOD 1
uses a table of values to find a positive integer that satisfies 𝑃3 (𝑛) = 351 (M1)
for example, a list showing at least 3 consecutive terms (… 325, 351, 378 … )

Note: Award (M1) for use of a GDC’s numerical solve or graph feature.

𝑛 = 26 (26th triangular number) A1

Note: Award A0 for 𝑛 = −27,26. Award A0 if additional solutions besides 𝑛 = 26 are given.
 METHOD 2
𝑛(𝑛+1)
attempts to solve 2 = 351 (𝑛 2 + 𝑛 − 702 = 0) for 𝑛 (M1)
−1±√12 −4(1)(−702)
𝑛= OR (𝑛 − 26)(𝑛 + 27) = 0
2
𝑛 = 26 (26th triangular number) A1

Note: Award A0 for 𝑛 = −27,26. Award A0 if additional solutions besides 𝑛 = 26 are given.

[2 marks]

(b.i) Show that 𝑃3 (𝑛) + 𝑃3 (𝑛 + 1) ≡ (𝑛 + 1)2.

[2]
Markscheme

attempts to form an expression for 𝑃3 (𝑛) + 𝑃3 (𝑛 + 1) in terms of 𝑛 M1

EITHER
𝑛(𝑛 + 1) (𝑛 + 1)(𝑛 + 2)
𝑃3 (𝑛) + 𝑃3 (𝑛 + 1) ≡ +
2 2
(𝑛+1)(2𝑛+2) 2(𝑛+1)(𝑛+1)
≡ (≡ ) A1
2 2

OR
𝑛2 𝑛 (𝑛 + 1)2 𝑛 + 1
𝑃3 (𝑛) + 𝑃3 (𝑛 + 1) ≡ ( + ) + ( + )
2 2 2 2
𝑛2 +𝑛 𝑛 2 +2𝑛+1+𝑛+1
≡( )+( ) (≡ 𝑛 2 + 2𝑛 + 1) A1
2 2

THEN
≡ (𝑛 + 1)2 AG

[2 marks]

(b.ii) State, in words, what the identity given in part (b)(i) shows for two consecutive triangular numbers.
[1]
Markscheme

the sum of the 𝑛th and (𝑛 + 1)th triangular numbers

is the (𝑛 + 1)th square number A1

[1 mark]

(b.iii) For 𝑛 = 4, sketch a diagram clearly showing your answer to part (b)(ii).
[1]
Markscheme
 A1

Note: Accept equivalent single diagrams, such as the one above, where the 4th and 5th triangular
numbers and the 5th square number are clearly shown.
Award A1 for a diagram that show 𝑃3 (4) (a triangle with 10 dots) and 𝑃3 (5) (a triangle with 15
dots) and 𝑃4 (5) (a square with 25 dots).

[1 mark]

(c) Show that 8𝑃3 (𝑛) + 1 is the square of an odd number for all 𝑛 ∈ ℤ+ .
[3]
Markscheme

METHOD 1
𝑛(𝑛+1)
8𝑃3 (𝑛) + 1 = 8 ( 2 ) + 1 (= 4𝑛(𝑛 + 1) + 1) A1
attempts to expand their expression for 8𝑃3(𝑛) + 1 (M1)
= 4𝑛 2 + 4𝑛 + 1
= (2𝑛 + 1)2 A1
and 2𝑛 + 1 is odd AG

METHOD 2
(𝑛+1)(𝑛+2)
8𝑃3 (𝑛) + 1 = 8((𝑛 + 1)2 − 𝑃3 (𝑛 + 1)) + 1 (= 8 ((𝑛 + 1)2 − ) + 1) A1
2
attempts to expand their expression for 8𝑃3(𝑛) + 1 (M1)
2 2 2
8(𝑛 + 2𝑛 + 1) − 4(𝑛 + 3𝑛 + 2) + 1 (= 4𝑛 + 4𝑛 + 1)
= (2𝑛 + 1)2 A1
and 2𝑛 + 1 is odd AG

Method 3
𝑛(𝑛+1)
8𝑃3 (𝑛) + 1 = 8 ( 2 ) + 1 (= (𝐴𝑛 + 𝐵)2 ) (where 𝐴, 𝐵 ∈ ℤ+ ) A1
attempts to expand their expression for 8𝑃3(𝑛) + 1 (M1)
4𝑛 2 + 4𝑛 + 1 (= 𝐴2 𝑛 2 + 2𝐴𝐵𝑛 + 𝐵 2 )
now equates coefficients and obtains 𝐵 = 1 and 𝐴 = 2
= (2𝑛 + 1)2 A1
and 2𝑛 + 1 is odd AG

[3 marks]

The 𝑛th pentagonal number can be represented by the arithmetic series

𝑃5 (𝑛) = 1 + 4 + 7 + ⋯ + (3𝑛 − 2).
𝑛(3𝑛−1)
(d) Hence show that 𝑃5 (𝑛) = for 𝑛 ∈ ℤ+ .
2
[3]
 Markscheme

EITHER
𝑢1 = 1 and 𝑑 = 3 (A1)
𝑛
substitutes their 𝑢1 and their 𝑑 into 𝑃5 (𝑛) = 2 (2𝑢1 + (𝑛 − 1)𝑑) M1
𝑛 𝑛
𝑃5 (𝑛) = 2 (2 + 3(𝑛 − 1)) (= 2 (2 + 3𝑛 − 3)) A1

OR
𝑢1 = 1 and 𝑢𝑛 = 3𝑛 − 2 (A1)
𝑛
substitutes their 𝑢1 and their 𝑢𝑛 into 𝑃5 (𝑛) = 2 (𝑢1 + 𝑢𝑛 ) M1
𝑛
𝑃5 (𝑛) = 2 (1 + 3𝑛 − 2) A1

OR
𝑃5 (𝑛) = (3(1) − 2) + (3(2) − 2) + (3(3) − 2) + ⋯ 3𝑛 − 2
𝑃5 (𝑛) = (3(1) + 3(2) + 3(3) + ⋯ + 3𝑛) − 2𝑛 (= 3(1 + 2 + 3 + ⋯ + 𝑛) − 2𝑛) (A1)
𝑛(𝑛+1)
substitutes 2 into their expression for 𝑃5 (𝑛) M1
𝑛(𝑛 + 1)
𝑃5 (𝑛) = 3 ( ) − 2𝑛
2
𝑛
𝑃5 (𝑛) = 2 (3(𝑛 + 1) − 4) A1

OR
attempts to find the arithmetic mean of 𝑛 terms (M1)
1+(3𝑛−2)
= A1
2
multiplies the above expression by the number of terms 𝑛
𝑛
𝑃5 (𝑛) = (1 + 3𝑛 − 2) A1
2

THEN
𝑛(3𝑛−1)
so 𝑃5 (𝑛) = AG
2

[3 marks]

(e) By using a suitable table of values or otherwise, determine the smallest positive integer, greater than 1, that
is both a triangular number and a pentagonal number.
[5]
Markscheme

METHOD 1
forms a table of 𝑃3 (𝑛) values that includes some values for 𝑛 > 5 (M1)
forms a table of 𝑃5 (𝑚) values that includes some values for 𝑚 > 5 (M1)

Note: Award (M1) if at least one 𝑃3 (𝑛) value is correct. Award (M1) if at least one 𝑃5 (𝑚) value
is correct. Accept as above for (𝑛 2 + 𝑛) values and (3𝑚2 − 𝑚) values.

𝑛 = 20 for triangular numbers (A1)

𝑚 = 12 for pentagonal numbers (A1)
Note: Award (A1) if 𝑛 = 20 is seen in or out of a table. Award (A1) if 𝑚 = 12 is seen in or out
of a table. Condone the use of the same parameter for triangular numbers and pentagonal
numbers, for example, 𝑛 = 20 for triangular numbers and 𝑛 = 12 for pentagonal numbers.

210 (is a triangular number and a pentagonal number) A1

Note: Award all five marks for 210 seen anywhere with or without working shown.

METHOD 2
EITHER
attempts to express 𝑃3 (𝑛) = 𝑃5 (𝑚) as a quadratic in 𝑛 (M1)
𝑛 2 + 𝑛 + (𝑚 − 3𝑚2 )(= 0) (or equivalent)
attempts to solve their quadratic in 𝑛 (M1)
2
−1 ± √12𝑚 − 4𝑚 + 1 −1 ± √12 − 4(𝑚 − 3𝑚2 )
𝑛= (= )
2 2

OR
attempts to express 𝑃3 (𝑛) = 𝑃5 (𝑚) as a quadratic in 𝑚 (M1)
2 2
3𝑚 − 𝑚 − (𝑛 + 𝑛)(= 0) (or equivalent)
attempts to solve their quadratic in 𝑚 (M1)
2
1 ± √12𝑛 − 12𝑛 + 1 1 ± √(−1)2 + 12(𝑛 2 + 𝑛)
𝑚= (= )
6 6

THEN
𝑛 = 20 for triangular numbers (A1)
𝑚 = 12 for pentagonal numbers (A1)
210 (is a triangular number and a pentagonal number) A1

METHOD 3
𝑛(𝑛 + 1) 𝑚(3𝑚 − 1)
=
2 2
let 𝑛 = 𝑚 + 𝑘 (𝑛 > 𝑚) and so 3𝑚2 − 𝑚 = (𝑚 + 𝑘)(𝑚 + 𝑘 + 1) M1
2𝑚2 − 2(𝑘 + 1)𝑚 − (𝑘 2 + 𝑘) = 0 A1
attempts to find the discriminant of their quadratic
and recognises that this must be a perfect square M1
2 2
Δ = 4(𝑘 + 1) + 8(𝑘 + 𝑘)
𝑁 2 = 4(𝑘 + 1)2 + 8(𝑘 2 + 𝑘) (= 4(𝑘 + 1)(3𝑘 + 1))
determines that 𝑘 = 8 leading to 2𝑚2 − 18𝑚 − 72 = 0 ⇒ 𝑚 = −3,12 and so 𝑚 = 12 A1
210 (is a triangular number and a pentagonal number) A1

METHOD 4
𝑛(𝑛 + 1) 𝑚(3𝑚 − 1)
=
2 2
let 𝑚 = 𝑛 − 𝑘 (𝑚 < 𝑛 ) and so 𝑛 2 + 𝑛 = (𝑛 − 𝑘)(3(𝑛 − 𝑘) − 1) M1
2𝑛 2 − 2(3𝑘 + 1)𝑛 + (3𝑘 2 + 𝑘) = 0 A1
attempts to find the discriminant of their quadratic
and recognises that this must be a perfect square M1
2 2
Δ = 4(3𝑘 + 1) − 8(3𝑘 + 𝑘)
𝑁 2 = 4(3𝑘 + 1)2 − 8(3𝑘 2 + 𝑘) (= 4(𝑘 + 1)(3𝑘 + 1))
 determines that 𝑘 = 8 leading to 2𝑛 2 − 50𝑛 + 200 = 0 ⇒ 𝑛 = 5,20 and so 𝑛 = 20 A1
210 (is a triangular number and a pentagonal number) A1

[5 marks]

(f) A polygonal number, 𝑃𝑟 (𝑛), can be represented by the series

𝑛
Σ (1 + (𝑚 − 1)(𝑟 − 2)) where 𝑟 ∈ ℤ+ , 𝑟 ≥ 3.
𝑚=1
(𝑟−2)𝑛2 −(𝑟−4)𝑛
Use mathematical induction to prove that 𝑃𝑟 (𝑛) = where 𝑛 ∈ ℤ+ .
2
[8]
Markscheme

Note: Award a maximum of R1M0M0A1M1A1A1R0 for a ‘correct’ proof using 𝑛 and 𝑛 + 1.

(𝑟−2)(12 )−(𝑟−4)(1)
consider 𝑛 = 1: 𝑃𝑟 (1) = 1 + (1 − 1)(𝑟 − 2) = 1 and 𝑃𝑟 (1) = 2
=1
so true for 𝑛 = 1 R1

(𝑟−2)(12 )−(𝑟−4)(1)
Note: Accept 𝑃𝑟 (1) = 1 and 𝑃𝑟 (1) = = 1.
2
Do not accept one-sided considerations such as '𝑃𝑟 (1) = 1 and so true for 𝑛 = 1'.
Subsequent marks after this R1 are independent of this mark can be awarded.

(𝑟−2)𝑘 2 −(𝑟−4)𝑘
Assume true for 𝑛 = 𝑘, ie. 𝑃𝑟 (𝑘) = M1
2

Note: Award M0 for statements such as “let 𝑛 = 𝑘 ”. The assumption of truth must be clear.
Subsequent marks after this M1 are independent of this mark and can be awarded.

Consider 𝑛 = 𝑘 + 1:
(𝑃𝑟 (𝑘 + 1) can be represented by the sum
𝑘+1 𝑘
Σ (1 + (𝑚 − 1)(𝑟 − 2)) = Σ (1 + (𝑚 − 1)(𝑟 − 2)) + (1 + 𝑘(𝑟 − 2)) and so
𝑚=1 𝑚=1
(𝑟−2)𝑘 2 −(𝑟−4)𝑘
𝑃𝑟 (𝑘 + 1) = + (1 + 𝑘(𝑟 − 2)) (𝑃𝑟 (𝑘 + 1) = 𝑃𝑟 (𝑘) + (1 + 𝑘(𝑟 − 2))) M1
2
(𝑟−2)𝑘 2 −(𝑟−4)𝑘+2+2𝑘(𝑟−2)
= A1
2
(𝑟−2)(𝑘 2 +2𝑘)−(𝑟−4)𝑘+2
= 2
(𝑟−2)(𝑘 2 +2𝑘+1)−(𝑟−2)−(𝑟−4)𝑘+2
= M1
2
(𝑟−2)(𝑘+1)2 −(𝑟−4)𝑘−(𝑟−4)
= (A1)
2
(𝑟−2)(𝑘+1)2 −(𝑟−4)(𝑘+1)
= A1
2
hence true for 𝑛 = 1 and 𝑛 = 𝑘 true ⇒ 𝑛 = 𝑘 + 1 true R1
therefore true for all 𝑛 ∈ ℤ+

Note: Only award the final R1 if the first five marks have been awarded. Award marks as
appropriate for solutions that expand both the LHS and (given) RHS of the equation.

[8 marks]

6. 20N.1.AHL.TZ0.H_5
 The first term in an arithmetic sequence is 4 and the fifth term is log2 625.
Find the common difference of the sequence, expressing your answer in the form log2 𝑝, where 𝑝 ∈ ℚ.
[5]
Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in
marking or structure.
𝑢5 = 4 + 4𝑑 = log2 625 (A1)
4𝑑 = log2 625 − 4 attempt to write an integer (eg 4 or 1) in terms of log2 M1 4𝑑 =
625
log2 625 − log2 16 attempt to combine two logs into one M1 4𝑑 = log2 ( 16 ) 𝑑 =
1
1 625 625 4 5
log2 ( 16 ) attempt to use power rule for logs M1 𝑑 = log2 ( 16 ) 𝑑 = log2 (2) A1
4
[5 marks]

Note: Award method marks in any order.

7. 20N.2.AHL.TZ0.H_11
A particle 𝑃 moves in a straight line such that after time 𝑡 seconds, its velocity, 𝑣 in m s−1, is given by 𝑣 =
𝜋
e−3𝑡 sin 6 𝑡, where 0 < 𝑡 < 2 .
(a) Find the times when 𝑃 comes to instantaneous rest.
[2]
Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in
marking or structure.
𝜋
(= 0.524) A1
6
𝜋
(= 1.05) A1
3
[2 marks]

At time 𝑡, 𝑃 has displacement 𝑠(𝑡); at time 𝑡 = 0, 𝑠(0) = 0.

(b) Find an expression for 𝑠 in terms of 𝑡.
[7]
Markscheme

attempt to use integration by parts M1 𝑠 = ∫ e−3𝑡 sin 6𝑡 d𝑡

EITHER
e−3𝑡 sin 6𝑡 e−3𝑡 sin 6𝑡 2e−3𝑡 cos 6𝑡
=− − ∫ − 2e−3𝑡 cos 6𝑡 d𝑡 A1 = − −( − ∫ − 4e−3𝑡 sin 6𝑡 d𝑡)
3 3 3
e−3𝑡 sin 6𝑡 2e−3𝑡 cos 6𝑡 -3e−3𝑡 sin 6𝑡−6e−3𝑡 cos 6𝑡
A1 = − −( + 4𝑠) 5𝑠 = M1
3 3 9
OR
e−3𝑡 cos 6𝑡 1 e−3𝑡 cos 6𝑡 e−3𝑡 sin 6𝑡 1
=− 6
− ∫ 2 e−3𝑡 cos 6𝑡 d𝑡 A1 = − 6
−( 12
+ ∫ 4 e−3𝑡 sin 6𝑡 d𝑡) A1
e−3𝑡 cos 6𝑡 e−3𝑡 sin 6𝑡 1 5 −2e−3𝑡 cos 6𝑡−e−3𝑡 sin 6𝑡
=− −( + 4 𝑠) 4 𝑠 = M1
6 12 12
THEN
e−3𝑡 ( sin 6𝑡+2 cos 6𝑡) 2 2
𝑠=− (+𝑐) A1 at 𝑡 = 0, 𝑠 = 0 ⇒ 0 = − 15 + 𝑐 M1 𝑐 = 15 A1 𝑠 =
15
 2 e−3𝑡 ( sin 6𝑡+2 cos 6𝑡)
−
15 15
[7 marks]

(c) Find the maximum displacement of 𝑃, in metres, from its initial position.
[2]
Markscheme
𝜋
−
𝜋 2 e 2 ( sin 𝜋+2 cos 𝜋)
EITHER substituting 𝑡 = 6 into their equation for 𝑠 (M1) (𝑠 = 15 − )
15

OR

using GDC to find maximum value (M1)

𝜋
OR evaluating ∫06 𝑣 d𝑡 (M1)
THEN
𝜋
2
= 0.161 (= 15 (1 + e−2 )) A1
[2 marks]

(d) Find the total distance travelled by 𝑃 in the first 1.5 seconds of its motion.
[2]
Markscheme

METHOD 1
1.5
EITHER distance required = ∫0 |e−3𝑡 sin 6𝑡| d𝑡 (M1)
𝜋 𝜋
1.5
OR distance required = ∫0 e−3𝑡 sin 6𝑡 d𝑡 + |∫ e−3𝑡 sin 6𝑡 d𝑡| + ∫𝜋 e−3𝑡 sin 6𝑡 d𝑡
6
𝜋
3
(M1)
6 3
(= 0.16105 … + 0.033479 … + 0.006806 … )
THEN
= 0.201 (m) A1 METHOD 2
using successive minimum and maximum values on the displacement graph (M1)
0.16105 … + (0.16105 … − 0.12757 … ) + (0.13453 … − 0.12757 … ) = 0.201 (m) A1
[2 marks]

At successive times when the acceleration of 𝑃 is 0 m s−2 , the velocities of 𝑃 form a geometric sequence. The
acceleration of 𝑃 is zero at times 𝑡1 , 𝑡2 , 𝑡3 where 𝑡1 < 𝑡2 < 𝑡3 and the respective velocities are 𝑣1 , 𝑣2 , 𝑣3 .
(e.i) Show that, at these times, tan 6𝑡 = 2.
[2]
Markscheme
d𝑣 d𝑣 d𝑣
valid attempt to find using product rule and set =0 M1 d𝑡 = e−3𝑡 6 cos 6𝑡 −
d𝑡 d𝑡
d𝑣
3e−3𝑡 sin 6𝑡 A1 d𝑡 = 0 ⇒ tan 6𝑡 = 2 AG
[2 marks]

𝜋
𝑣 𝑣
(e.ii) Hence show that 𝑣2 = 𝑣3 = −e−2 .
1 2
 [5]
Markscheme
1
attempt to evaluate 𝑡1 , 𝑡2 , 𝑡3 in exact form M1 6𝑡1 = arctan 2 (⇒ 𝑡1 = 6 arctan 2) 6𝑡2 =
𝜋 1 𝜋 1
𝜋 + arctan 2 (⇒ 𝑡2 = 6 + 6 arctan 2) 6𝑡3 = 2𝜋 + arctan 2 (⇒ 𝑡3 = 3 + 6 arctan 2) A1
Note: The A1 is for any two consecutive correct, or showing that 6𝑡2 = 𝜋 + 6𝑡1 or 6𝑡3 = 𝜋 +
6𝑡2.
2
showing that sin 6𝑡𝑛+1 = −sin 6𝑡𝑛 eg tan 6𝑡 = 2 ⇒ sin 6𝑡 = ± M1A1 showing
√5
𝜋 𝜋 𝜋
e−3𝑡𝑛+1 − −3( +𝑘) −3𝑘 −
that e−3𝑡𝑛 = e M1 eg e 2 ÷e =e 6 2

Note: Award the A1 for any two consecutive terms.

𝜋
𝑣3 𝑣
= 𝑣2 = −e−2 AG
𝑣2 1
[5 marks]

8. 20N.2.AHL.TZ0.H_11
A particle 𝑃 moves in a straight line such that after time 𝑡 seconds, its velocity, 𝑣 in m s−1, is given by 𝑣 =
𝜋
e−3𝑡 sin 6 𝑡, where 0 < 𝑡 < 2 .
(a) Find the times when 𝑃 comes to instantaneous rest.
[2]
Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in
marking or structure.
𝜋
(= 0.524) A1
6
𝜋
(= 1.05) A1
3
[2 marks]

At time 𝑡, 𝑃 has displacement 𝑠(𝑡); at time 𝑡 = 0, 𝑠(0) = 0.

(b) Find an expression for 𝑠 in terms of 𝑡.
[7]
Markscheme

attempt to use integration by parts M1 𝑠 = ∫ e−3𝑡 sin 6𝑡 d𝑡

EITHER
e−3𝑡 sin 6𝑡 e−3𝑡 sin 6𝑡 2e−3𝑡 cos 6𝑡
=− − ∫ − 2e−3𝑡 cos 6𝑡 d𝑡 A1 = − −( − ∫ − 4e−3𝑡 sin 6𝑡 d𝑡)
3 3 3
e−3𝑡 sin 6𝑡 2e−3𝑡 cos 6𝑡 -3e−3𝑡 sin 6𝑡−6e−3𝑡 cos 6𝑡
A1 = − −( + 4𝑠) 5𝑠 = M1
3 3 9
OR
e−3𝑡 cos 6𝑡 1 e−3𝑡 cos 6𝑡 e−3𝑡 sin 6𝑡 1
=− − ∫ 2 e−3𝑡 cos 6𝑡 d𝑡 A1 = − −( + ∫ 4 e−3𝑡 sin 6𝑡 d𝑡) A1
6 6 12
e−3𝑡 cos 6𝑡 e−3𝑡 sin 6𝑡 1 5 −2e−3𝑡 cos 6𝑡−e−3𝑡 sin 6𝑡
=− −( + 𝑠) 𝑠 = M1
6 12 4 4 12
THEN
e−3𝑡 ( sin 6𝑡+2 cos 6𝑡) 2 2
𝑠=− (+𝑐) A1 at 𝑡 = 0, 𝑠 = 0 ⇒ 0 = − 15 + 𝑐 M1 𝑐 = 15 A1 𝑠 =
15
2 e−3𝑡 ( sin 6𝑡+2 cos 6𝑡)
−
15 15
[7 marks]
(c) Find the maximum displacement of 𝑃, in metres, from its initial position.
[2]
Markscheme
𝜋
−
𝜋 2 e 2 ( sin 𝜋+2 cos 𝜋)
EITHER substituting 𝑡 = 6 into their equation for 𝑠 (M1) (𝑠 = 15 − )
15

OR

using GDC to find maximum value (M1)

𝜋
OR evaluating ∫06 𝑣 d𝑡 (M1)
THEN
𝜋
2
= 0.161 (= 15 (1 + e−2 )) A1
[2 marks]

(d) Find the total distance travelled by 𝑃 in the first 1.5 seconds of its motion.
[2]
Markscheme

METHOD 1
1.5
EITHER distance required = ∫0 |e−3𝑡 sin 6𝑡| d𝑡 (M1)
𝜋 𝜋
1.5
OR distance required = ∫0 e−3𝑡 sin 6𝑡 d𝑡 + |∫ e−3𝑡 sin 6𝑡 d𝑡| + ∫𝜋 e−3𝑡 sin 6𝑡 d𝑡
6
𝜋
3
(M1)
6 3
(= 0.16105 … + 0.033479 … + 0.006806 … )
THEN
= 0.201 (m) A1 METHOD 2
using successive minimum and maximum values on the displacement graph (M1)
0.16105 … + (0.16105 … − 0.12757 … ) + (0.13453 … − 0.12757 … ) = 0.201 (m) A1
[2 marks]

At successive times when the acceleration of 𝑃 is 0 m s−2 , the velocities of 𝑃 form a geometric sequence. The
acceleration of 𝑃 is zero at times 𝑡1 , 𝑡2 , 𝑡3 where 𝑡1 < 𝑡2 < 𝑡3 and the respective velocities are 𝑣1 , 𝑣2 , 𝑣3 .
(e.i) Show that, at these times, tan 6𝑡 = 2.
[2]
Markscheme
d𝑣 d𝑣 d𝑣
valid attempt to find using product rule and set =0 M1 d𝑡 = e−3𝑡 6 cos 6𝑡 −
d𝑡 d𝑡
d𝑣
3e−3𝑡 sin 6𝑡 A1 d𝑡 = 0 ⇒ tan 6𝑡 = 2 AG
[2 marks]

𝜋
𝑣 𝑣
(e.ii) Hence show that 𝑣2 = 𝑣3 = −e−2 .
1 2
[5]
Markscheme
 1
attempt to evaluate 𝑡1 , 𝑡2 , 𝑡3 in exact form M1 6𝑡1 = arctan 2 (⇒ 𝑡1 = 6 arctan 2) 6𝑡2 =
𝜋 1 𝜋 1
𝜋 + arctan 2 (⇒ 𝑡2 = 6 + 6 arctan 2) 6𝑡3 = 2𝜋 + arctan 2 (⇒ 𝑡3 = 3 + 6 arctan 2) A1
Note: The A1 is for any two consecutive correct, or showing that 6𝑡2 = 𝜋 + 6𝑡1 or 6𝑡3 = 𝜋 +
6𝑡2.
2
showing that sin 6𝑡𝑛+1 = −sin 6𝑡𝑛 eg tan 6𝑡 = 2 ⇒ sin 6𝑡 = ± M1A1 showing
√5
𝜋 𝜋 𝜋
e−3𝑡𝑛+1 − −3( +𝑘) −3𝑘 −
that e−3𝑡𝑛 = e 2M1 eg e ÷e
6 =e 2

Note: Award the A1 for any two consecutive terms.

𝜋
𝑣3 𝑣
= 𝑣2 = −e−2 AG
𝑣2 1
[5 marks]

9. 19N.2.AHL.TZ0.H_1
A geometric sequence has 𝑢4 = −70 and 𝑢7 = 8.75. Find the second term of the sequence.
[5]
Markscheme
8.75
𝑢1 𝑟 3 = −70, 𝑢1 𝑟 6 = 8.75 (M1) 𝑟 3 = −70 = −0.125 (A1) ⇒ 𝑟 = −0.5 (A1) valid
−70
attempt to find 𝑢2 (M1) for example: 𝑢1 = −0.125 = 560 𝑢2 = 560 × −0.5 = −280
A1 [5 marks]

10. 19M.2.AHL.TZ1.H_11
Consider the equation 𝑥 5 − 3𝑥 4 + 𝑚𝑥 3 + 𝑛𝑥 2 + 𝑝𝑥 + 𝑞 = 0, where 𝑚, 𝑛, 𝑝, 𝑞 ∈ ℝ.
The equation has three distinct real roots which can be written as log2 𝑎, log2 𝑏 and log2 𝑐.
The equation also has two imaginary roots, one of which is 𝑑i where 𝑑 ∈ ℝ.
(a) Show that 𝑎𝑏𝑐 = 8.
[5]
Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in
marking or structure.
recognition of the other root = −𝑑i (A1) log2 𝑎 + log2 𝑏 + log2 𝑐 + 𝑑i − 𝑑i = 3 M1A1
Note: Award M1 for sum of the roots, A1 for 3. Award A0M1A0 for just log2 𝑎 + log2 𝑏 +
log2 𝑐 = 3. log2 𝑎𝑏𝑐 = 3 (M1) ⇒ 𝑎𝑏𝑐 = 23 A1 𝑎𝑏𝑐 = 8 AG [5 marks]

The values 𝑎, 𝑏, and 𝑐 are consecutive terms in a geometric sequence.

(b) Show that one of the real roots is equal to 1.
[3]
Markscheme

METHOD 1 let the geometric series be 𝑢1, 𝑢1 𝑟, 𝑢1 𝑟 2 (𝑢1 𝑟)3 = 8 M1 𝑢1 𝑟 = 2 A1 hence

𝑏 𝑐 2
one of the roots is log2 2 = 1 R1 METHOD 2 𝑎 = 𝑏 𝑏 = 𝑎𝑐 ⇒ 𝑏3 = 𝑎𝑏𝑐 = 8 M1 𝑏 =
2 A1 hence one of the roots is log2 2 = 1 R1 [3 marks]

(c) Given that 𝑞 = 8𝑑 2, find the other two real roots.

[9]
 Markscheme

METHOD 1 product of the roots is 𝑟1 × 𝑟2 × 1 × 𝑑i × −𝑑i = −8𝑑 2 (M1)(A1) 𝑟1 × 𝑟2 =

−8 A1 sum of the roots is 𝑟1 + 𝑟2 + 1 + 𝑑i + −𝑑i = 3 (M1)(A1) 𝑟1 + 𝑟2 = 2 A1
solving simultaneously (M1) 𝑟1 = −2, 𝑟2 = 4 A1A1 METHOD 2 product of the
roots log2 𝑎 × log2 𝑏 × log2 𝑐 × 𝑑i × −𝑑i = −8𝑑 2 M1A1 log2 𝑎 × log2 𝑏 × log2 𝑐 = −8
2 2
A1 EITHER 𝑎, 𝑏, 𝑐 can be written as 𝑟 , 2, 2𝑟 M1 (log2 𝑟 ) (log2 2)(log2 2𝑟) = −8 attempt
1
to solve M1 (1 − log2 𝑟)(1 + log2 𝑟) = −8 log2 𝑟 = ±3 𝑟 = 8 , 8 A1A1 OR 𝑎, 𝑏, 𝑐 can
4 4 1
be written as 𝑎, 2, 𝑎 M1 (log2 𝑎)(log2 2) (log2 𝑎) = −8 attempt to solve M1 𝑎 = 4 , 16
1
A1A1 THEN 𝑎 and 𝑐 are , 16 (A1) roots are −2, 4 A1 [9 marks]
4

11. 18N.2.AHL.TZ0.H_1
Consider a geometric sequence with a first term of 4 and a fourth term of −2.916.
(a) Find the common ratio of this sequence.
[3]
Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in
marking or structure.
𝑢4 = 𝑢1 𝑟3 ⇒ −2.916 = 4𝑟 3 (A1)
solving, 𝑟 = −0.9 (M1)A1
[3 marks]

(b) Find the sum to infinity of this sequence.

[2]
Markscheme
4 40
𝑆∞ = 1−(−9) (M1) = 19 (= 2.11) A1 [2 marks]

12. 18M.2.AHL.TZ1.H_7
It is known that the number of fish in a given lake will decrease by 7% each year unless some new fish are
added. At the end of each year, 250 new fish are added to the lake.
At the start of 2018, there are 2500 fish in the lake.
(a) Show that there will be approximately 2645 fish in the lake at the start of 2020.
[3]
Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in
marking or structure.
EITHER
2019: 2500 × 0.93 + 250 = 2575 (M1)A1 2020: 2575 × 0.93 + 250 M1 OR 2020: 2500
× 0.932 + 250(0.93 + 1) M1M1A1 Note: Award M1 for starting with 2500, M1 for
multiplying by 0.93 and adding 250 twice. A1 for correct expression. Can be shown in recursive
form. THEN (= 2644.75) = 2645 AG [3 marks]

(b) Find the approximate number of fish in the lake at the start of 2042.
 [5]
Markscheme

2020: 2500 × 0.932 + 250(0.93 + 1)

2042: 2500 × 0.9324 + 250(0.9323 + 0.9322 + … + 1) (M1)(A1)
(0.9324−1)
= 2500 × 0.9324 + 250 (0.93−1)
(M1)(A1)
=3384 A1 Note: If recursive formula used, award M1 for un = 0.93 un−1 and u0 or u1 seen (can
be awarded if seen in part (a)). Then award M1A1 for attempt to find u24 or u25 respectively
(different term if other than 2500 used) (M1A0 if incorrect term is being found) and A2 for
correct answer. Note: Accept all answers that round to 3380. [5 marks]

13. 18M.2.AHL.TZ1.H_1
The 3rd term of an arithmetic sequence is 1407 and the 10th term is 1183.
(b) Calculate the number of positive terms in the sequence.
[3]
Markscheme

1471 + (n − 1)(−32) > 0 (M1)

1471
⇒ n < 32 + 1
n < 46.96… (A1) so 46 positive terms A1 [3 marks]

14. EXN.1.SL.TZ0.4
The first three terms of an arithmetic sequence are 𝑢1 , 5𝑢1 − 8 and 3𝑢1 + 8.
(a) Show that 𝑢1 = 4.
[2]
Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid
teachers in preparing for external assessment in the new MAA course. There may be minor
differences in formatting compared to formal exam papers.
EITHER uses 𝑢2 − 𝑢1 = 𝑢3 − 𝑢2 (M1) (5𝑢1 − 8) − 𝑢1 = (3𝑢1 + 8) − (5𝑢1 − 8) 6𝑢1 =
𝑢 +𝑢3 𝑢1 +(3𝑢1 +8)
24 A1 OR uses 𝑢2 = 1 2 (M1) 5𝑢1 − 8 = 3𝑢1 = 12 A1 THEN
2
so 𝑢1 = 4 AG [2 marks]

(b) Prove that the sum of the first 𝑛 terms of this arithmetic sequence is a square number.
[4]
Markscheme
𝑛 𝑛
𝑑=8 (A1) uses 𝑆𝑛 = 2 (2𝑢1 + (𝑛 − 1)𝑑) M1 𝑆𝑛 = 2 (8 + 8(𝑛 − 1)) A1 = 4𝑛 2 =
(2𝑛)2 A1 Note: The final A1 can be awarded for clearly explaining that 4𝑛 2 is a square
number. so sum of the first 𝑛 terms is a square number AG [4 marks]

15. EXN.2.SL.TZ0.7
Helen and Jane both commence new jobs each starting on an annual salary of $70,000. At the start of each new
year, Helen receives an annual salary increase of $2400.
Let $𝐻𝑛 represent Helen’s annual salary at the start of her 𝑛th year of employment.
(a) Show that 𝐻𝑛 = 2400𝑛 + 67 600.
 [2]
Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid
teachers in preparing for external assessment in the new MAA course. There may be minor
differences in formatting compared to formal exam papers.
uses 𝐻𝑛 = 𝐻1 + (𝑛 − 1)𝑑 with 𝐻1 = 70 000 and 𝑑 = 2400 (M1) 𝐻𝑛 = 70 000 +
2400(𝑛 − 1) A1 so 𝐻𝑛 = 2400𝑛 + 67 600 AG [2 marks]

At the start of each new year, Jane receives an annual salary increase of 3% of her previous year’s annual
salary.
Jane’s annual salary, $𝐽𝑛 , at the start of her 𝑛th year of employment is given by 𝐽𝑛 = 70 000(1.03)𝑛−1.
(b) Given that 𝐽𝑛 follows a geometric sequence, state the value of the common ratio, 𝑟.
[1]
Markscheme

𝑟 = 1.03 A1 [1 mark]

At the start of year 𝑁, Jane’s annual salary exceeds Helen’s annual salary for the first time.
(c.i) Find the value of 𝑁.
[3]
Markscheme

evidence of use of an appropriate table or graph or GDC numerical solve feature to find the value
of 𝑁 such that 𝐽𝑛 > 𝐻𝑛 (M1) EITHER for example, an excerpt from an appropriate table

(A1) OR for example, use of a GDC numerical solve feature to obtain 𝑁 = 10.800 … (A1)
Note: Award A1 for an appropriate graph. Condone use of a continuous graph. THEN 𝑁 =
11 A1 [3 marks]

(c.ii) For the value of 𝑁 found in part (c) (i), state Helen’s annual salary and Jane’s annual salary, correct to the
nearest dollar.
[2]
Markscheme

𝐻11 = 94 000 ($) A1 𝐽11 = 94 074 ($) A1 Helen’s annual salary is $94 000 and Jane’s
annual salary is $94 074 Note: Award A1 for a correct 𝐻11 value and A1 for a correct 𝐽11 value
seen in part (c) (i). [2 marks]

(d) Find Jane’s total earnings at the start of her 10th year of employment. Give your answer correct to the
nearest dollar.
[4]
Markscheme
 at the start of the 10th year, Jane will have worked for 9 years so the value of 𝑆9 is required
𝐽1 (𝑟𝑛 −1)
R1 Note: Award R1 if 𝑆9 is seen anywhere. uses 𝑆𝑛 = with 𝐽1 = 70 000, 𝑟 =
𝑟−1
70 000((1.03)9−1)
1.03 and 𝑛 = 9 (M1) Note: Award M1 if 𝑛 = 10 is used. 𝑆9 = =
1.03−1
711 137.42 … (A1) = 711 137 ($) Jane’s total earnings are $711 137 (correct to the nearest
dollar) [4 marks]

16. 23N.1.SL.TZ1.4
The sum of the first 𝑛 terms of an arithmetic sequence is given by 𝑆𝑛 = 𝑝𝑛 2 − 𝑞𝑛, where 𝑝 and 𝑞 are positive
constants.
It is given that 𝑆5 = 65 and 𝑆6 = 96.
[N/A]
[[N/A]]
(a) Find the value of 𝑝 and the value of q.
[5]
Markscheme

METHOD 1 attempt to form at least one equation, using either 𝑆5 or 𝑆6 (M1) 65 =

25𝑝 − 5𝑞 (13 = 5𝑝 − 𝑞) and 96 = 36𝑝 − 6𝑞 (16 = 6𝑝 − 𝑞) (A1) valid attempt
to solve simultaneous linear equations in p and q and by substituting or eliminating one of the
variables. (M1) 𝑝 = 3 , 𝑞 = 2 A1A1 Note: If candidate does not
explicitly state their values of p and q, but gives 𝑆𝑛 = 3𝑛 2 − 2𝑛 , award final two marks as
A1A0. METHOD 2 attempt to form at least one equation, using either 𝑆5 or 𝑆6
5
(M1) 65 = 2 (2𝑢1 + 4𝑑) (26 = 2𝑢1 + 4𝑑) and 96 = 3(2𝑢1 + 5𝑑) (32 = 2𝑢1 + 5𝑑)
(A1) valid attempt to solve simultaneous linear equations in 𝑢1 and 𝑑 by substituting or
eliminating one of the variables. (M1) 𝑢1 = 1, 𝑑 = 6 A1 𝑆𝑛 =
𝑛 2
(2 + 6(𝑛 − 1)) = 3𝑛 − 2𝑛 𝑝 = 3 and 𝑞 = 2 A1 Note: If candidate does not
2
explicitly state their values of p and q, do not award the final mark.
[5 marks]

(b) Find the value of 𝑢6.

[2]
Markscheme

𝑢6 = 𝑆6 − 𝑆5 OR substituting their values of 𝑢1 and 𝑑 into 𝑢6 = 𝑢1 + 5𝑑 OR substituting their

6
value of 𝑢1 into 96 = 2 (𝑢1 + 𝑢6 ) (M1) (𝑢6 =)96 − 65 OR (𝑢6 =)1 + 5 × 6
OR 96 = 3(1 + 𝑢6 ) = 31 A1
[2 marks]

17. 23N.2.SL.TZ1.8
Give your answers to parts (a)(ii), (c)(i) and (d) correct to two decimal places.
Daniela and Sorin have each recently received some money. Daniela won a cash prize and Sorin received an
inheritance.
Daniela had two options to choose from to receive her winnings. In both options she receives a payment on the
first day of each month for three years.
Option A Each payment is $4200.
Option B The first payment is $1500. In each month which follows, the payment is 4% more than the previous
month.
[N/A]
[[N/A]]
(a) Find the total amount Daniela would receive if she chooses
[[N/A]]
(a.i) Option A;
[2]
Markscheme

4200 × 36 (A1) = 151200 = ($)151000 A1

[2 marks]

(a.ii) Option B.
[3]
Markscheme

Note: The first time an answer is not given to two decimal places in parts (a)(ii), (c)(i) or (d), the
final A1 in that part is not awarded. recognizing sum of a geometric sequence is required
1500(1−1.0436 )
(M1) (A1) = 116397.4707 … = ($)116397.47 A1
1−1.04
[3 marks]

Sorin received an inheritance of $160 000. Sorin invested his inheritance in an account that pays a nominal
annual interest rate of 5% per annum, compounded monthly. The interest is added on the last day of each
month.
(b) Write down an expression for the value of Sorin’s investment after 𝑛 years.
[1]
Markscheme

5 12𝑛
Sorin’s future value after 𝑛 years = 160000 (1 + 100×12) A1
[1 mark]

Daniela chose Option B and received her first payment on 1st January 2023. Sorin invested his inheritance on
the same day.
(c.i) Find the total value of Daniela’s winnings and Sorin’s investment on the last day of the sixth month.
[3]
Markscheme

Note: The first time an answer is not given to two decimal places in parts (a)(ii), (c)(i) or (d), the
5 6
final A1 in that part is not awarded. Sorin’s total 160000 (1 + 100×12) (164041.89 … )
1500(1−1.046 )
(A1) Daniela’s total = (= 9949.46 ⋯ ) (A1) total value =
1−1.04
($)173991.36 A1
[3 marks]

(c.ii) Find the minimum number of complete months before the total value of Daniela’s winnings and Sorin’s
investment is at least $257 000.
[3]
 Markscheme

5 𝑚 1500(1−1.04𝑚 )
EITHER (finding number of months, 𝑚) 160000 (1 + 100×12) + (≥ 257000)
1−1.04
(A1) 𝑚 ≥ 28.4412 … OR (𝑚 = 28 ⇒) 254707 AND (𝑚 = 29 ⇒) 259954
(A1) Note: Condone use of an equation or strict inequality. OR (finding number of years, 𝑛)
5 12×𝑛 1500(1−1.0412×𝑛 )
160000 (1 + 100×12) + (≥ 257000) (A1) 𝑛 ≥
1−1.04
2.37010 …(years) (A1) Note: Condone use of an equation or strict inequality. THEN
𝑚 = 29 (months) A1
[3 marks]

At the end of the three years, Daniela invested $30 000 for a further six years in a second account that pays a
nominal interest rate of 𝑟% per annum compounded quarterly.
(d) Find the value of 𝑟 if this investment grows to $41 000 after six years.
[3]
Markscheme

Note: The first time an answer is not given to two decimal places in parts (a)(ii), (c)(i) or (d), the
final A1 in that part is not awarded. EITHER 𝑁 = 24 𝑃𝑉 = ∓30000 𝑃𝑀𝑇 = 0 𝐹𝑉 = ∓41000
𝑃/𝑌 = 4 𝐶/𝑌 = 4 OR 𝑁 = 6 𝑃𝑉 = ∓30000 𝑃𝑀𝑇 = 0 𝐹𝑉 = ±41000 𝑃/𝑌 = 1 𝐶/𝑌 = 4
(M1)(A1)
Note: Award (M1) for an attempt to use a financial app in their technology with at least two
entries seen, award (A1) for all entries correct. PV and FV must have opposite signs.
OR
𝑟 6×4
30000 (1 + 100×4) = 41000 (M1)(A1)
Note: Award (M1) for attempting to substitute into compound interest formula, award (A1) for
correct equation.
THEN
5.24027 …
(𝑟 =)5.24% A1
[3 marks]

18. 23N.2.SL.TZ2.8
Give your answers to parts (a)(ii), (c)(i) and (d) correct to two decimal places.
Daniela and Sorin have each recently received some money. Daniela won a cash prize and Sorin received an
inheritance.
Daniela had two options to choose from to receive her winnings. In both options she receives a payment on the
first day of each month for three years.
Option A Each payment is $5500.
Option B The first payment is $2000 . In each month which follows, the payment is 6% more than the
previous month.
(a) Find the total amount Daniela would receive if she chooses
[[N/A]]
(a.i) Option A;
[2]
Markscheme

5500 × 36 (A1) = ($)198000 A1

 [2 marks]

(a.ii) Option B.
[3]
Markscheme

Note: The first time an answer is not given to two decimal places in parts (a)(ii), (c)(i) or (d), the
final A1 in that part is not awarded. recognizing sum of a geometric sequence is required
2000(1 − 1.0636 )
(M1) (A1) = 238241.7333 … = ($)238241.73 A1
1 − 1.06
[3 marks]

Sorin received an inheritance of $120 000. Sorin invested his inheritance in an account that pays a nominal
annual interest rate of 4% per annum, compounded monthly. The interest is added on the last day of each
month.
(b) Write down an expression for the value of Sorin’s investment after 𝑛 years.
[1]
Markscheme

4 12𝑛
Sorin’s future value after 𝑛 years = 120000 (1 + ) A1
100 × 12
[1 mark]

Daniela chose Option B and received her first payment on 1st January 2023. Sorin invested his inheritance on
the same day.
(c.i) Find the total value of Daniela’s winnings and Sorin’s investment on the last day of the sixth month.
[3]
Markscheme

Note: The first time an answer is not given to two decimal places in parts (a)(ii), (c)(i) or (d), the
4 6
final A1 in that part is not awarded. Sorin’s total = 120000 (1 + ) (= 122420.09)
100 × 12
2000(1 − 1.066 )
(A1) Daniela’s total = (= 13950.64) (A1) total value =
1 − 1.06
($)136370.73 A1
[3 marks]

(c.ii) Find the minimum number of complete months before the total value of Daniela’s winnings and Sorin’s
investment is at least $250 000.
[3]
Markscheme

4 𝑚 2000(1 − 1.06𝑚)
EITHER (finding number of months, 𝑚) 120000 (1 + 100 × 12) + (≥
1 − 1.06
250000) (A1) 𝑚 ≥ 26.0905 OR (𝑚 = 26 ⇒ ) 249157 … AND (𝑚 =
27 ⇒) 258692 … (A1) Note: Condone use of an equation or strict inequality.
4 12 × 𝑛 2000(1 − 1.0612 × 𝑛 )
OR (finding number of years, 𝑛) 120000 (1 + 100 × 12) + (≥
1 − 1.06
250000) (A1) 𝑛 ≥ 2.17421 …(years) (A1) Note: Condone use of an
equation or strict inequality. THEN 𝑚 = 27(𝑚𝑜𝑛𝑡ℎ𝑠) A1
 [3 marks]

At the end of the three years, Daniela invested $40 000 for a further six years in a second account that pays a
nominal interest rate of 𝑟% per annum compounded quarterly.
(d) Find the value of 𝑟 if this investment grows to $53 000 after six years.
[3]
Markscheme

Note: The first time an answer is not given to two decimal places in parts (a)(ii), (c)(i) or (d), the
final A1 in that part is not awarded. EITHER 𝑁 = 24 𝑃𝑉 = ∓40000 𝑃𝑀𝑇 = 0 𝐹𝑉 = ±53000
𝑃/𝑌 = 4 𝐶/𝑌 = 4 OR 𝑁 = 6 𝑃𝑉 = ∓40000 𝑃𝑀𝑇 = 0 𝐹𝑉 = ±530000 𝑃/𝑌 = 1 𝐶/𝑌 = 4
(M1)(A1)
Note: Award (M1) for an attempt to use a financial app in their technology with at least two
entries seen, and award (A1) for all entries correct. PV and FV must have opposite signs.
OR
𝑟 6×4
40000 (1 + 100 × 4) = 53000 (M1)(A1)
Note: Award (M1) for attempting to substitute into compound interest formula, award (A1) for
correct equation.
THEN
4.71781 …
(𝑟 =)4.72% A1
[3 marks]

19. 23M.1.SL.TZ1.8
Consider the arithmetic sequence 𝑢1, 𝑢2, 𝑢3, … .
The sum of the first 𝑛 terms of this sequence is given by 𝑆𝑛 = 𝑛 2 + 4𝑛.
[N/A]
[[N/A]]
(a.i) Find the sum of the first five terms.
[2]
Markscheme

recognition that 𝑛 = 5 (M1) 𝑆5 = 45 A1 [2 marks]

(a.ii) Given that 𝑆6 = 60, find 𝑢6.

[2]
Markscheme

METHOD 1 recognition that 𝑆5 + 𝑢6 = 𝑆6 (M1) 𝑢6 = 15 A1 METHOD 2

6
recognition that 60 = 2 (𝑆1 + 𝑢6 ) (M1) 60 = 3(5 + 𝑢6 ) 𝑢6 = 15 A1 METHOD 3
substituting their 𝑢1 and 𝑑 values into 𝑢1 + (𝑛 − 1)𝑑 (M1)
𝑢6 = 15 A1

[2 marks]

(b) Find 𝑢1.

[2]
 Markscheme

recognition that 𝑢1 = 𝑆1 (may be seen in (a)) OR substituting their 𝑢6 into 𝑆6 (M1) OR

6
equations for 𝑆5 and 𝑆6 in terms of 𝑢1 and 𝑑 1 + 4 OR 60 = 2 (𝑈1 + 15) 𝑢1 = 5 A1 [2
marks]

(c) Hence or otherwise, write an expression for 𝑢𝑛 in terms of 𝑛.

[3]
Markscheme

EITHER valid attempt to find 𝑑 (may be seen in (a) or (b)) (M1) 𝑑 = 2 (A1) OR
2 2
valid attempt to find 𝑆𝑛 − 𝑆𝑛−1 (M1) 𝑛 + 4𝑛 − (𝑛 − 2𝑛 + 1 + 4𝑛 − 4) (A1) OR
2 𝑛
equating 𝑛 + 4𝑛 = 2 (5 + 𝑢𝑛 ) (M1) 2𝑛 + 8 = 5 + 𝑢𝑛 (or equivalent) (A1) THEN
𝑢𝑛 = 5 + 2(𝑛 − 1) OR 𝑢𝑛 = 2𝑛 + 3 A1

[3 marks]

Consider a geometric sequence, 𝑣𝑛 , where 𝑣2 = 𝑢1 and 𝑣4 = 𝑢6.

(d) Find the possible values of the common ratio, 𝑟.
[3]
Markscheme

recognition that 𝑣2 𝑟 2 = 𝑣4 OR (𝑣3 )2 = 𝑣2 × 𝑣4 (M1) 𝑟 2 = 3 OR 𝑣3 = (±)5√3

(A1) 𝑟 = ±√3 A1 Note: If no working shown, award M1A1A0 for √3. [3 marks]

(e) Given that 𝑣99 < 0, find 𝑣5.

[2]
Markscheme
45
recognition that 𝑟 is negative (M1) 𝑣5 = −15√3 (= − ) A1 [2 marks]
√3

20. 23M.1.SL.TZ2.2
Consider an arithmetic sequence with 𝑢1 = 0.6 and 𝑢4 = 0.15.
[N/A]
[[N/A]]
(a) Find the common difference, 𝑑.
[2]
Markscheme

𝑢1 + 3𝑑 = 𝑢4 (M1) 0.6 + 3𝑑 = 0.15 𝑑 = −0.15 A1 [2 marks]

The following table shows the probability distribution of a discrete random variable 𝑋 such that 𝑃(𝑋 = 𝑛) =
𝑢𝑛
, where 𝑛 ∈ ℤ+ , 1 ≤ 𝑛 ≤ 4 and 𝑘 ∈ ℝ+ .
𝑘
 𝑛 1 2 3 4
𝑃(𝑋 = 𝑛) 0.6 𝑢2 𝑢3 0.15
𝑘 𝑘 𝑘 𝑘
(b) Find the value of 𝑘.
[4]
Markscheme

METHOD 1 𝑢2 = 0.45 or 𝑢3 = 0.3 (may be seen in their equation) (A1) summing their
0.6 𝑢2 𝑢3 0.15 0.6 0.45 0.3 0.15
probabilities to 1 (seen anywhere) (M1) 𝑘 + 𝑘 + 𝑘 + 𝑘 = 1 𝑘 + 𝑘 + 𝑘 + 𝑘 =
1.5
1 (or equivalent) (A1) = 1 𝑘 = 1.5 A1 METHOD 2 (using 𝑆𝑛 formula) 𝑆4 =
𝑘
4 4 0.6 −0.15 4
(2(0.6) + (4 − 1)(−0.15)) OR 𝑆4 = 2 (2 ( 𝑘 ) + (4 − 1) ( )) OR 𝑆4 = 2 (0.6 + 0.15)
2 𝑘
4 0.6 0.15
OR 𝑆4 = 2 ( 𝑘 + ) (or equivalent) (A1) summing their probabilities to 1 (seen
𝑘
𝑢1 𝑢2 𝑢3 𝑢4
anywhere) (M1) + + + = 1 OR 𝑢1 + 𝑢2 + 𝑢3 + 𝑢4 = 𝑘 OR 𝑆4 = 1
𝑘 𝑘 𝑘 𝑘
4
OR 𝑆4 = 𝑘 2 (2(0.6) + (4 − 1)(−0.15)) = 𝑘 (or equivalent) (A1) 𝑘 = 1.5 A1 [4
marks]

21. 23M.1.AHL.TZ1.10
Consider the arithmetic sequence 𝑢1, 𝑢2, 𝑢3, … .
The sum of the first 𝑛 terms of this sequence is given by 𝑆𝑛 = 𝑛 2 + 4𝑛.
[N/A]
[[N/A]]
(a.i) Find the sum of the first five terms.
[2]
Markscheme

recognition that 𝑛 = 5 (M1) 𝑆5 = 45 A1 [2 marks]

(a.ii) Given that 𝑆6 = 60, find 𝑢6.

[2]
Markscheme

METHOD 1 recognition that 𝑆5 + 𝑢6 = 𝑆6 (M1) 𝑢6 = 15 A1 METHOD 2

6
recognition that 60 = 2 (𝑆1 + 𝑢6 ) (M1) 60 = 3(5 + 𝑢6 ) 𝑢6 = 15 A1 METHOD 3
substituting their 𝑢1 and 𝑑 values into 𝑢1 + (𝑛 − 1)𝑑 (M1)
𝑢6 = 15 A1

[2 marks]

(b) Find 𝑢1.

[2]
Markscheme
 recognition that 𝑢1 = 𝑆1 (may be seen in (a)) OR substituting their 𝑢6 into 𝑆6 (M1) OR
6
equations for 𝑆5 and 𝑆6 in terms of 𝑢1 and 𝑑 1 + 4 OR 60 = 2 (𝑈1 + 15) 𝑢1 = 5 A1 [2
marks]

(c) Hence or otherwise, write an expression for 𝑢𝑛 in terms of 𝑛.

[3]
Markscheme

EITHER valid attempt to find 𝑑 (may be seen in (a) or (b)) (M1) 𝑑 = 2 (A1) OR
valid attempt to find 𝑆𝑛 − 𝑆𝑛−1 (M1) 𝑛 2 + 4𝑛 − (𝑛 2 − 2𝑛 + 1 + 4𝑛 − 4) (A1) OR
𝑛
equating 𝑛 2 + 4𝑛 = 2 (5 + 𝑢𝑛 ) (M1) 2𝑛 + 8 = 5 + 𝑢𝑛 (or equivalent) (A1) THEN
𝑢𝑛 = 5 + 2(𝑛 − 1) OR 𝑢𝑛 = 2𝑛 + 3 A1

[3 marks]

Consider a geometric sequence, 𝑣𝑛 , where 𝑣2 = 𝑢1 and 𝑣4 = 𝑢6.

(d) Find the possible values of the common ratio, 𝑟.
[3]
Markscheme

recognition that 𝑣2 𝑟 2 = 𝑣4 OR (𝑣3 )2 = 𝑣2 × 𝑣4 (M1) 𝑟 2 = 3 OR 𝑣3 = (±)5√3

(A1) 𝑟 = ±√3 A1 Note: If no working shown, award M1A1A0 for √3. [3 marks]

(e) Given that 𝑣99 < 0, find 𝑣5.

[2]
Markscheme
45
recognition that 𝑟 is negative (M1) 𝑣5 = −15√3 (= − ) A1 [2 marks]
√3

22. 23M.1.AHL.TZ2.12
(a) By using an appropriate substitution, show that ∫ cos√𝑥𝑑𝑥 = 2√𝑥sin√𝑥 + 2 cos√𝑥 + 𝐶.
[6]
Markscheme

let 𝑡 = √𝑥 M1 𝑡 2 = 𝑥 ⇒ 2𝑡 𝑑𝑡 = 𝑑𝑥 A1 so ∫ cos√𝑥𝑑𝑥 = 2∫ 𝑡 cos 𝑡 𝑑𝑡 A1

attempts integration by parts (M1) 𝑢 = 2𝑡, 𝑑𝑣 = cos 𝑡 𝑑𝑡, 𝑑𝑢 = 2 𝑑𝑡, 𝑣 = sin 𝑡
2∫ 𝑡 cos 𝑡 𝑑𝑡 = 2𝑡 sin 𝑡 − 2∫ sin 𝑡 𝑑𝑡 (A1) = 2𝑡 sin 𝑡 + 2 cos 𝑡 + 𝐶 A1 substitution of
𝑡 = √𝑥 ⇒ ∫ cos√𝑥𝑑𝑥 = 2√𝑥sin√𝑥 + 2 cos√𝑥 + 𝐶 AG [6 marks]

The following diagram shows part of the curve 𝑦 = cos√𝑥 for 𝑥 ≥ 0.

The curve intersects the 𝑥-axis at 𝑥1, 𝑥2, 𝑥3, 𝑥4 ….
(2𝑛−1)2 𝜋2
The 𝑛th 𝑥-intercept of the curve, 𝑥𝑛 , is given by 𝑥𝑛 = , where 𝑛 ∈ ℤ+ .
4
(b) Write down a similar expression for 𝑥𝑛+1.
[1]
Markscheme

(2(𝑛+1)−1)2 𝜋2 (2𝑛+1)2 𝜋2
𝑥𝑛+1 = (= ) A1 [1 mark]
4 4

The regions bounded by the curve and the 𝑥-axis are denoted by 𝑅1, 𝑅2, 𝑅3, … , as shown on the above
diagram.
(c) Calculate the area of region 𝑅𝑛 .
Give your answer in the form 𝑘𝑛𝜋, where 𝑘 ∈ ℤ+ .
[7]
Markscheme
𝑥
area of 𝑅𝑛 is |∫𝑥 𝑛+1 cos √𝑥 𝑑𝑥| (M1) Note: Modulus may be seen at a later stage. =
𝑛
(2𝑛+1)2 𝜋2

|[2√𝑥sin√𝑥 + 2 cos√𝑥](2𝑛−1)
4
2 𝜋2 | A1 Note: Condone +𝐶 at this stage. attempts to
4
|2𝑛+1|𝜋 |2𝑛+1|𝜋
substitute their limits into their integrated expression (M1) = 2 | × sin +
2 2
|2𝑛+1|𝜋 |2𝑛−1|𝜋 |2𝑛−1|𝜋 |2𝑛−1|𝜋 𝑛 (2𝑛+1)𝜋
cos −| × sin + cos || A1 = 2 |(−1) −
2 2 2 2 2
(2𝑛−1)𝜋 (2𝑛+1)𝜋 (2𝑛−1)𝜋
|(−1)𝑛+1 2 || (or equivalent) A1 = 2 |(−1)𝑛 + (−1) 𝑛
| A1 =
2 2
𝑛 4𝑛𝜋
2 |(−1) 2 | = 4𝑛𝜋 A1
Note: Award a maximum of (M1)A1M1A1A1A0A0 for only attempting to calculate
𝑥𝑛+1
∫𝑥 cos √𝑥 𝑑𝑥, and not applying the modulus. [7 marks]
𝑛

(d) Hence, show that the areas of the regions bounded by the curve and the 𝑥-axis, 𝑅1, 𝑅2, 𝑅3, … , form an
arithmetic sequence.
 [3]
Markscheme

EITHER attempts to find (𝑑 =)𝑅𝑛+1 − 𝑅𝑛 M1 (𝑑 =)4(𝑛 + 1)𝜋 − 4𝑛𝜋 = 4𝜋 A1

Note: Award M0 for consideration of special cases for example 𝑅3 and 𝑅2. Accept 𝑑 = 𝑘𝜋.
which is a constant (common difference is 4𝜋) R1 OR an arithmetic sequence is of the
form 𝑢𝑛 = 𝑑𝑛 + 𝑐 (𝑢𝑛 = 𝑑𝑛 + 𝑢1 − 𝑑) M1 attempts to compare 𝑢𝑛 = 𝑑𝑛 + 𝑐 (𝑢𝑛 =
𝑑𝑛 + 𝑢1 − 𝑑) and 𝑅𝑛 = 4𝑛𝜋 M1 𝑑 = 4𝜋 and 𝑐 = 0 (𝑢1 − 𝑑 = 0) A1 Note: Accept
𝑑 = 𝑘𝜋. THEN so the areas of the regions form an arithmetic sequence AG [3 marks]

23. 23M.3.AHL.TZ2.2
This question asks you to examine linear and quadratic functions constructed in systematic ways using
arithmetic sequences.
Consider the function 𝐿(𝑥) = 𝑚𝑥 + 𝑐 for 𝑥 ∈ ℝ where 𝑚, 𝑐 ∈ ℝ and 𝑚, 𝑐 ≠ 0.
Let 𝑟 ∈ ℝ be the root of 𝐿(𝑥) = 0.
If 𝑚, 𝑟 and 𝑐, in that order, are in arithmetic sequence then 𝐿(𝑥) is said to be an AS-linear function.
[N/A]
[[N/A]]
(a) Show that 𝐿(𝑥) = 2𝑥 − 1 is an AS-linear function.
[2]
Markscheme
1 1 1 1 3
𝑚 = 2, 𝑐 = −1 𝑟 = 2 (A1) 2, 2, −1 EITHER 𝑑 (= 2 − 2 = −1 − 2) = − 2 A1 OR
3
this sequence has a common difference of − 2 A1 OR the (arithmetic) mean of 2 and −1 is
1
A1 THEN hence 𝐿(𝑥) = 2𝑥 − 1 is an AS-linear function AG [2 marks]
2

Consider 𝐿(𝑥) = 𝑚𝑥 + 𝑐.
𝑐
(b.i) Show that 𝑟 = − 𝑚.
[1]
Markscheme
𝑐
(𝐿(𝑟) = 0 ⇒)𝑚𝑟 + 𝑐 = 0 A1 𝑟 = − 𝑚 AG Note: Award A0 for numerical
verification from 𝐿(𝑥) = 2𝑥 − 1 in part (a). [1 mark]

𝑚2
(b.ii) Given that 𝐿(𝑥) is an AS-linear function, show that 𝐿(𝑥) = 𝑚𝑥 − 𝑚+2.
[4]
Markscheme

METHOD 1 EITHER attempts to use (𝑑 =)𝑟 − 𝑚 = 𝑐 − 𝑟 (M1) Note: Award (M1) for
𝑐 𝑐
attempting to use (𝑑 =)𝑚 − 𝑟 = 𝑟 − 𝑐. (𝑑 =) − 𝑚 − 𝑚 = 𝑐 − (− 𝑚) A1 Note: Award
𝑐 𝑐−𝑚
A1 for (𝑑 =) − 𝑚 − 𝑚 = 2
. removes the denominator 𝑚 from their expression involving 𝑚
𝑚+𝑐
and 𝑐 (M1) 𝑚2 + 𝑐𝑚 + 2𝑐 = 0 (or equivalent) OR attempts to use =𝑟 (M1)
2
2𝑐
𝑚+𝑐 =−𝑚 A1 removes the denominator 𝑚 from their expression involving 𝑚 and
2
𝑐 (M1) 𝑚 + 𝑐𝑚 + 2𝑐 = 0 (or equivalent) OR attempts to use 𝑐 = 𝑚 + 2𝑑 (M1)
 𝑐 𝑐
𝑐 = 𝑚 + 2 (− 𝑚 − 𝑚) A1 Note: Award A1 for 𝑐 = 𝑚 + 2 (𝑐 − (− 𝑚)). removes the
denominator 𝑚 from their expression involving 𝑚 and 𝑐 (M1) 𝑚2 + 𝑐𝑚 + 2𝑐 = 0 (or
equivalent) OR attempts to use 𝑟 = 𝑚 + 𝑑 and 𝑐 = 𝑚 + 2𝑑 (𝑐 = 𝑚 + 2(𝑟 − 𝑚)) (M1)
2 𝑐−𝑚
𝑚 + 𝑑𝑚 + 𝑚 + 2𝑑 = 0 (or equivalent) A1 substitutes 𝑑 = 2 into their expression
involving 𝑚 and 𝑑 (M1) 𝑚 + 𝑐𝑚 + 2𝑐 = 0 (or equivalent) THEN 𝑐(𝑚 + 2) = −𝑚2 ⇒
2
𝑚2 𝑚2
𝑐 = − 𝑚+2 A1 Note: Award A1 for a convincing demonstration that 𝑐 = − 𝑚+2. so
𝑚2
𝐿(𝑥) = 𝑚𝑥 − 𝑚+2 AG Note: Do not accept working backwards from the AG. METHOD
2 considers 𝐿(𝑥) = 𝑚𝑥 − 𝑚𝑟 attempts to use (𝑑 =)𝑟 − 𝑚 = 𝑐 − 𝑟 (M1) Note: Award
(M1) for attempting to use (𝑑 =)𝑚 − 𝑟 = 𝑟 − 𝑐. (𝑑 =)𝑟 − 𝑚 = −𝑚𝑟 − 𝑟 A1 attempts to
𝑚 𝑚2
express 𝑟 in terms of 𝑚 (M1) 2𝑟 + 𝑚𝑟 = 𝑚 ⇒ 𝑟 = 𝑚+2 A1 so 𝐿(𝑥) = 𝑚𝑥 − 𝑚+2
AG Note: Do not accept working backwards from the AG. [4 marks]

(b.iii) State any further restrictions on the value of 𝑚.

[1]
Markscheme

𝑚 ≠ −2 (𝑚 ≠ 0) A1 [1 mark]

There are only three integer sets of values of 𝑚, 𝑟 and 𝑐, that form an AS-linear function. One of these is
𝐿(𝑥) = −𝑥 − 1.
(c) Use part (b) to determine the other two AS-linear functions with integer values of 𝑚, 𝑟 and 𝑐.
[3]
Markscheme

attempts to find an integer value of 𝑚 (M1) e.g. uses the result that 𝑚 + 2 exactly divides 2
OR uses a table OR uses a graph and slider OR uses systematic trial and error Note: Award
𝑚2
(M1) for solving 𝑚2 = 𝑘(𝑚 + 2) for 𝑚 or solving 𝑚𝑟 − 𝑚+2 = 0 for 𝑚 or solving 𝑚2 + 𝑐𝑚 +
2𝑐 = 0 for 𝑚. 𝑚 = −4 OR 𝑚 = −3 (A1) −4, 2, 8 OR −3, 3, 9 𝐿(𝑥) = −4𝑥 +
8, 𝐿(𝑥) = −3𝑥 + 9 A1 Note: Award (M1)(A1)A0 for −4𝑥 + 8 and −3𝑥 + 9. [3 marks]

Consider the function 𝑄(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 for 𝑥 ∈ ℝ where 𝑎 ∈ ℝ, 𝑎 ≠ 0 and 𝑏, 𝑐 ∈ ℝ.

Let 𝑟1, 𝑟2 ∈ ℝ be the roots of 𝑄(𝑥) = 0.
(d) Write down an expression for
[[N/A]]
(d.i) the sum of roots, 𝑟1 + 𝑟2, in terms of 𝑎 and 𝑏.
[1]
Markscheme
𝑏
−𝑎 A1 [1 mark]

(d.ii) the product of roots, 𝑟1 𝑟2, in terms of 𝑎 and 𝑐.

[1]
Markscheme
 𝑐
A1 [1 mark]
𝑎

If 𝑎, 𝑟1, 𝑏, 𝑟2 and 𝑐, in that order, are in arithmetic sequence, then 𝑄(𝑥) is said to be an AS-quadratic function.

(e) Given that 𝑄(𝑥) is an AS-quadratic function,

[[N/A]]
(e.i) write down an expression for 𝑟2 − 𝑟1 in terms of 𝑎 and 𝑏;
[1]
Markscheme

𝑏−𝑎 A1 Note: Award marks as appropriate in parts (e) (ii) and (iii) for use of 𝑟1 − 𝑟𝑏 =
𝑎 − 𝑏. [1 mark]

𝑎2 −𝑎𝑏−𝑏
(e.ii) use your answers to parts (d)(i) and (e)(i) to show that 𝑟1 = .
2𝑎
[2]
Markscheme

𝑏 𝑎2 −𝑎𝑏−𝑏
attempts to eliminate 𝑟2 M1 2𝑟1 = − 𝑎 − (𝑏 − 𝑎) ⇒ 2𝑟1 = (or equivalent)
𝑎
𝑎2 −𝑎𝑏−𝑏
A1 Note: Award A1 for a correct alternative form of ±𝑟1 or ±2𝑟1. so 𝑟1 = AG
2𝑎
Note: Do not accept working backwards from the AG. [2 marks]

1
(e.iii) use the result from part (e)(ii) to show that 𝑏 = 0 or 𝑎 = − 2.
[3]
Markscheme
𝑎+𝑏
METHOD 1 EITHER (𝑟1 =) (A1) attempts to equate two expressions for either 𝑟1 or
2
𝑎+𝑏 𝑎2 −𝑎𝑏−𝑏 𝑎2 −𝑎𝑏−𝑏
2𝑟1 in terms of 𝑎 and 𝑏 M1 = OR 𝑎 + 𝑏 = OR 𝑏 − 𝑟1 = 𝑟1 −
2 2𝑎 𝑎
𝑎2 −𝑎𝑏−𝑏 𝑎2 −𝑎𝑏−𝑏
𝑎 (A1) attempts to use 𝑏 − 𝑟1 = 𝑟1 − 𝑎 with 𝑟1 = M1 𝑏 − ( )=
2𝑎 2𝑎
𝑎2 −𝑎𝑏−𝑏 𝑎2 −𝑎𝑏−𝑏
2𝑎
− 𝑎 OR (𝑟1 =)𝑎 + 𝑑 (A1) attempts to use 𝑟1 = 𝑎 + 𝑑 with 𝑟1 = 2𝑎
and 𝑑 =
𝑏−𝑎 𝑎2 −𝑎𝑏−𝑏 𝑏−𝑎
M1 2𝑎 = 𝑎 + 2 THEN 2𝑎2 + 2𝑎𝑏 = 2𝑎2 − 2𝑎𝑏 − 2𝑏 OR 𝑎 + 𝑎𝑏 = 𝑎2 − 2
2
𝑎𝑏 − 𝑏 4𝑎𝑏 + 2𝑏 = 0 OR 2𝑎𝑏 + 𝑏 = 0 2𝑏(2𝑎 + 1) = 0 OR 𝑏(2𝑎 + 1) = 0 A1 Note:
Award (A1)M1 for any valid approach that correctly leads to 2𝑎𝑏 + 𝑏 = 0 (or equivalent).
1
Do not accept numerical verification from the AG. so 𝑏 = 0 or 𝑎 = − 2 AG METHOD
2 (𝑏 =)𝑎 + 2𝑑 OR (𝑟1 =)𝑎 + 𝑑 (A1) attempts to equate two expressions for either 𝑟1 or
𝑎2 −𝑎(𝑎+2𝑑)−(𝑎+2𝑑)
2𝑟1 in terms of 𝑎 and 𝑑 M1 𝑎 + 𝑑 = OR 2(𝑎 + 𝑑) =
2𝑎
𝑎2 −𝑎(𝑎+2𝑑)−(𝑎+2𝑑)
2𝑎2 + 4𝑎𝑑 + 𝑎 + 2𝑑 = 0 (2𝑎 + 1)(𝑎 + 2𝑑) = 0 A1 Note: Do not
𝑎
1
accept numerical verification from the AG. so 𝑏 = 0 or 𝑎 = − 2 AG [3 marks]

Consider the case where 𝑏 = 0.

(f) Determine the two AS-quadratic functions that satisfy this condition.
[5]
 Markscheme
𝑎 𝑎 𝑎
METHOD 1 𝑟1 = 2 OR 𝑟2 = − 2 OR 𝑑 = − 2 (A1) 𝑐 = −𝑎 (A1) attempts to find
𝑎
the values of 𝑎 (M1) EITHER the roots of 𝑎𝑥 2 − 𝑎 = 0 are ±1 and 2 = ±1 OR
𝑎 𝑎3 𝑐 𝑎2 𝑎3
substitutes 𝑥 = ± into 𝑎𝑥 2 − 𝑎 = 0 giving − 𝑎 = 0 OR (𝑟1 𝑟2 =) = − ⇒𝑐= and
2 4 𝑎 4 4
𝑎3 𝑎3
so −𝑎 = − ⇒ − 𝑎 = 0 Note: Award (M1) for attempting to find the values of 𝑎 from their
4 4
𝑎3 𝑎 𝑎 𝑎3
arithmetic sequence expressed in terms of 𝑎. OR 𝑐 − 𝑟2 = 𝑟2 − 𝑏 ⇒ − 4 − (− 2) = − 2 ⇒ 4 −
𝑎 = 0 THEN 𝑎 = ±2 (A1) (𝑟1 = ±1, 𝑏 = 0, 𝑟2 = ∓1, 𝑐 = ∓2) so 𝑄(𝑥) = 2𝑥 2 −
2, 𝑄(𝑥) = −2𝑥 2 + 2 A1 Note: Award A0 for 2𝑥 2 − 2 and −2𝑥 2 + 2.
Award (A1)(A1)(M1)(A0)A0 for obtaining either 𝑎 = 2 or 𝑎 = −2. METHOD 2 𝑟1 = −𝑑 OR
𝑟2 = 𝑑 OR 𝑎 = −2𝑑 (A1) 𝑐 = 2𝑑 (A1) attempts to find the values of 𝑑 (M1)
EITHER the roots of −2𝑑𝑥 2 + 2𝑑 = 0 are ±1 OR substitutes 𝑥 = ±𝑑 into −2𝑑𝑥 2 + 2𝑑 = 0
𝑐 2𝑑
giving −2𝑑 3 + 2𝑑 = 0 OR attempts to use 𝑟1 𝑟2 = 𝑎 to form −𝑑 2 = −2𝑑 Note: Award (M1)
for attempting to find the values of 𝑑 from their arithmetic sequence expressed in terms of 𝑑.
THEN 𝑑 = ±1 (A1) (𝑎 = ±2, 𝑟1 = ±1, 𝑏 = 0, 𝑟2 = ∓1, 𝑐 = ∓2) so 𝑄(𝑥) = 2𝑥 2 −
2
2, 𝑄(𝑥) = −2𝑥 + 2 A1 Note: Award A0 for 2𝑥 2 − 2 and −2𝑥 2 + 2.
Award (A1)(A1)(M1)(A0)A0 for obtaining either 𝑑 = 1 or 𝑑 = −1. METHOD 3 𝑎 = 2𝑟1 OR
𝑟2 = −𝑟1 OR 𝑑 = −𝑟1 (A1) 𝑐 = −2𝑟1 (A1) attempts to find the values of 𝑟1
(M1) EITHER the roots of 2𝑟1 𝑥 − 2𝑟1 = 0 are ±1 OR substitutes 𝑥 = ±𝑟1 into 2𝑟1 𝑥 2 − 2𝑟1 =
2
𝑐 −2𝑟
0 giving 2𝑟1 3 − 2𝑟1 = 0 OR attempts to use 𝑟1 𝑟2 = 𝑎 to form −𝑟1 2 = 2𝑟 1 Note: Award (M1)
1
for attempting to find the values of 𝑟1 from their arithmetic sequence expressed in terms of 𝑟1.
THEN 𝑟1 = ±1 (A1) (𝑎 = ±2, 𝑟1 = ±1, 𝑏 = 0, 𝑟2 = ∓1, 𝑐 = ∓2) so 𝑄(𝑥) = 2𝑥 2 −
2, 𝑄(𝑥) = −2𝑥 2 + 2 A1 Note: Award A0 for 2𝑥 2 − 2 and −2𝑥 2 + 2.
Award (A1)(A1)(M1)(A0)A0 for obtaining either 𝑟1 = 1 or 𝑟1 = −1. [5 marks]

1
Now consider the case where 𝑎 = − 2.
(g.i) Find an expression for 𝑟1 in terms of 𝑏.
[2]
Markscheme
1
attempts to express 𝑟1 in terms of 𝑏 with 𝑎 = − 2. (M1) Note: Do not award (M1) if 𝑎 =
1 𝑎+𝑏 𝑎2 −𝑎𝑏−𝑏
is used. EITHER uses 𝑟1 = OR uses 𝑟1 = OR uses 𝑟1 − 𝑎 = 𝑏 − 𝑟1 THEN
2 2 2𝑎
1
2𝑏−1 𝑏 1 𝑏−
2
𝑟1 = 4
(= 2 − 4 , = 2
) A1 [2 marks]

(g.ii) Hence or otherwise, determine the exact values of 𝑏 and 𝑐 such that AS-quadratic functions are formed.
−𝑝±𝑞√𝑠
Give your answers in the form where 𝑝, 𝑞, 𝑠 ∈ ℤ+.
2
[5]
Markscheme
1
METHOD 1 EITHER substitutes their expression for 𝑟1 with 𝑎 = − 2 into 𝑄(𝑥)(= 0)
2𝑏−1 1 2𝑏−1 2 2𝑏−1 6𝑏+1 3𝑏 1
(M1) 𝑄 ( ) (= 0) ⇒ − 2 ( ) +𝑏( ) + 𝑐(= 0) OR 𝑟2 = (= + 4)
4 4 4 4 2
 1 6𝑏+1
substitutes their expression for 𝑟2 with 𝑎 = − 2 into 𝑄(𝑥)(= 0) (M1) 𝑄 ( ) (= 0) ⇒
4
1 6𝑏+1 2 6𝑏+1 4𝑏+1 1
−2( ) +𝑏( ) + 𝑐(= 0) THEN 𝑐 = (= 2𝑏 + 2) (seen anywhere) A1
4 4 2
−5±2√5
4𝑏2 + 20𝑏 + 5 = 0 attempts to solve their quadratic in 𝑏 (M1) 𝑏 = A1
2
4𝑏+1 −9±4√5
substitutes into 𝑐 = 𝑐= A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as
2 2
decimal values. Note: Award a maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i).
1
METHOD 2 substitutes their expressions for 𝑟1 and 𝑟2 with 𝑎 = − 2 into 𝑄(𝑥) (M1)
1 2𝑏−1 6𝑏+1 1 3 1 1 4𝑏+1 1
− 2 (𝑥 − ( )) (𝑥 − ( )) − 2 𝑥 2 + 𝑏𝑥 − 8 𝑏2 + 8 𝑏 + 32 𝑐 = (= 2𝑏 + 2) (seen
4 4 1
1 3 1 1
anywhere) A1 2𝑏 + = − 𝑏2 + 𝑏 + 4𝑏2 + 20𝑏 + 5 = 0 attempts to solve their
2 8 8 32
−5±2√5 4𝑏+1 −9±4√5
quadratic in 𝑏 (M1) 𝑏 = 2 A1 substitutes into 𝑐 = 2 𝑐 = 2 A1
Note: Award A0A0 for 𝑏 and 𝑐 expressed as decimal values. Note: Award a maximum of
6𝑏+1 3𝑏 1
(M1)A1(M1)A0A0FT for FT from part (g) (i). METHOD 3 𝑟2 = 4 (= 2 + 4) substitutes
1 𝑐 2𝑏−1 6𝑏+1 𝑐
their expressions for 𝑟1 and 𝑟2 with 𝑎 = − into 𝑟1 𝑟2 = (M1) ( )( )= 1 (or
2 𝑎 4 4 −2
4𝑏+1 1
equivalent) EITHER 𝑐 = (= 2𝑏 + 2) (seen anywhere) A1
2
1 2𝑏−1 6𝑏+1 6𝑏+1 2𝑏−1 1
OR − 2 ( )( )−( )= − (− 2) A1
4 4 4 4
−5±2√5
THEN 4𝑏2 + 20𝑏 + 5 = 0 attempts to solve their quadratic in 𝑏 (M1) 𝑏 = 2
4𝑏+1 −9±4√5
A1 substitutes into 𝑐 = 2 𝑐 = 2 A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as
decimal values. Note: Award a maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i).
1 −𝑏±√𝑏2 +2𝑐
METHOD 4 attempts to equate two expressions for 𝑟1 with 𝑎 = − 2 (M1) =
−1
2𝑏−1 2𝑏+1 4𝑏+1 1
(±√𝑏2 + 2𝑐 = )𝑐= (= 2𝑏 + 2) (seen anywhere) A1 12𝑏2 − 4𝑏 − 1 +
4 4 2
1
32 (2𝑏 + 2) = 0 (4𝑏2 + 20𝑏 + 5 = 0) attempts to solve their quadratic in 𝑏 (M1) 𝑏 =
−5±2√5 4𝑏+1 −9±4√5
A1 substitutes into 𝑐 = 2 𝑐 = 2 A1 Note: Award A0A0 for 𝑏 and 𝑐
2
expressed as decimal values. Note: Award a maximum of (M1)A1(M1)A0A0FT for FT from
1
part (g) (i). METHOD 5 EITHER 𝑟1 = 𝑑 − 2 substitutes their expression for 𝑟1 in terms of
1 1 1 1 2 1
𝑑with 𝑎 = − 2 into 𝑄(𝑥)(= 0) (M1) 𝑄 (𝑑 − 2) (= 0) ⇒ − 2 (𝑑 − 2) + 𝑏 (𝑑 − 2) +
1 1
𝑐(= 0) OR 𝑟2 = 3𝑑 − 2 substitutes their expression for 𝑟2 in terms of 𝑑 with 𝑎 = − 2 into
1 1 1 2 1
𝑄(𝑥)(= 0) (M1) 𝑄 (3𝑑 − 2) (= 0) ⇒ − 2 (3𝑑 − 2) + 𝑏 (3𝑑 − 2) + 𝑐(= 0) THEN 𝑏 =
1 1
2𝑑 − 2 and 𝑐 = 4𝑑 − 2 (seen anywhere) A1 4𝑑 2 + 8𝑑 − 1 = 0 attempts to solve their
−2±√5 −5±2√5 4𝑏+1
quadratic in 𝑑 (M1) 𝑑 = 𝑏= A1 substitutes into 𝑐 = 𝑐=
2 2 2
−9±4√5
A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as decimal values. Note: Award a
2
1
maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i). METHOD 6 𝑟1 = 𝑑 − 2 and 𝑟2 =
1 1 𝑐
3𝑑 − substitutes their expression for 𝑟1 and 𝑟2 in terms of 𝑑 with 𝑎 = − into 𝑟1 𝑟2 =
2 2 𝑎
1 1 𝑐 1
(M1) (𝑑 − 2) (3𝑑 − 2) = 1 (or equivalent) 𝑐 = 4𝑑 − 2 4𝑑 2 + 8𝑑 − 1 = 0 attempts to solve
−2
−2±√5 −5±2√5 4𝑏+1
their quadratic in 𝑑 (M1) 𝑑 = 𝑏= A1 substitutes into 𝑐 = 𝑐=
2 2 2
 −9±4√5
A1 Note: Award A0A0 for 𝑏 and 𝑐 expressed as decimal values. Note: Award a
2
maximum of (M1)A1(M1)A0A0FT for FT from part (g) (i). [5 marks]

24. 22N.2.SL.TZ0.4
geometric sequence has a first term of 50 and a fourth term of 86.4.
The sum of the first 𝑛 terms of the sequence is 𝑆𝑛 .
Find the smallest value of 𝑛 such that 𝑆𝑛 > 33 500.
[5]
Markscheme

86.4 = 50𝑟 3 (A1)

3 86.4
𝑟 = 1.2 (= √ 50 ) seen anywhere (A1)
50(1.2𝑛−1)
> 33500 OR 250(1.2𝑛 − 1) = 33500 (A1)
0.2
attempt to solve their geometric 𝑆𝑛 inequality or equation (M1)
sketch OR 𝑛 > 26.9045, 𝑛 = 26.9 OR 𝑆26 = 28368.8 OR 𝑆27 = 34092.6 OR algebraic
manipulation involving logarithms
𝑛 = 27 accept 𝑛 ≥ 27 A1

[5 marks]

25. 22N.3.AHL.TZ0.1
In this question you will investigate series of the form
𝑛
𝛴 𝑖 𝑞 = 1𝑞 + 2𝑞 + 3𝑞 + ⋯ + 𝑛 𝑞 where 𝑛, 𝑞 ∈ ℤ+
𝑖=1
and use various methods to find polynomials, in terms of 𝐧, for such series.
When 𝑞 = 1, the above series is arithmetic.
𝑛 1
(a) Show that 𝛴 𝑖 = 2 𝑛(𝑛 + 1).
𝑖=1
[1]
Markscheme

EITHER
𝑛
𝑆𝑛 = (2 × 1 + (𝑛 − 1) × 1)
2
A1

OR
𝑢1 = 1 and either 𝑢𝑛 = 𝑛 or 𝑑 = 1 stated explicitly A1

OR
1 + 2 + ⋯ + 𝑛 (or equivalent) stated explicitly A1

THEN
𝑛
𝑆𝑛 = (1 + 𝑛) AG
2

Note: Award A0 for a numerical verification.

[1 mark]
Consider the case when 𝑞 = 2.
𝑛
The following table gives values of 𝑛 2 and 𝛴 𝑖 2 for 𝑛 = 1, 2, 3.
𝑖=1

(b.i) Write down the value of 𝑝.

[1]
Markscheme

14 A1

[1 mark]

(b.ii) The sum of the first 𝑛 square numbers can be expressed as a cubic polynomial with three terms:
𝑛
𝛴 𝑖 2 = 𝑎1 𝑛 + 𝑎2 𝑛2 + 𝑎3 𝑛 3 where 𝑎1 , 𝑎2 , 𝑎3 ∈ ℚ+ .
𝑖=1
Hence, write down a system of three linear equations in 𝑎1 , 𝑎2 and 𝑎3.
[3]
Markscheme

𝑎1 + 𝑎2 + 𝑎3 = 1 A1
2𝑎1 + 4𝑎2 + 8𝑎3 = 5 A1
3𝑎1 + 9𝑎2 + 27𝑎3 = 14 A1

Note: For the third A mark, award A1FT for 3𝑎1 + 9𝑎2 + 27𝑎3 = 𝑝 where 𝑝 is their answer to
part (b) (i).

[3 marks]

(b.iii) Hence, find the values of 𝑎1 , 𝑎2 and 𝑎3.

[2]
Markscheme
1 1
𝑎1 = 6 (= 0.166666 … ≈ 0.167), 𝑎2 = 2 (= 0.5),
1
𝑎3 = 3 (= 0.333333 … ≈ 0.333) A2

Note: Award A1 if only two of 𝑎1 , 𝑎2 , 𝑎3 are correct.

Only award FT if three linear equations, each in 𝑎1 , 𝑎2 and 𝑎3 are stated in part (b) (ii) or (iii).
Award A2FT for their 𝑎1 , 𝑎2 and 𝑎3.
Award A1FT for their 𝑎1 , 𝑎2 and 𝑎3 = 0.
 [2 marks]

You will now consider a method that can be generalized for all values of 𝑞.
Consider the function 𝑓(𝑥) = 1 + 𝑥 + 𝑥 2 + ⋯ + 𝑥 𝑛 , 𝑛 ∈ ℤ+.
(c) Show that 𝑥𝑓′(𝑥) = 𝑥 + 2𝑥 2 + 3𝑥 3 + ⋯ + 𝑛𝑥 𝑛 .
[1]
Markscheme

𝑓′(𝑥) = 1 + 2𝑥 + 3𝑥 2 + ⋯ + 𝑛𝑥 𝑛−1 A1
𝑛
Note: Award A1 for 𝑓′(𝑥) = 𝛴 𝑖𝑥 𝑖−1.
𝑖=1

⇒ 𝑥𝑓′(𝑥) = 𝑥 + 2𝑥 2 + 3𝑥 3 + ⋯ + 𝑛𝑥 𝑛 AG

[1 mark]

Let 𝑓1 (𝑥) = 𝑥𝑓′(𝑥) and consider the following family of functions:

𝑓2 (𝑥) = 𝑥𝑓1 ′(𝑥)
𝑓3 (𝑥) = 𝑥𝑓2 ′(𝑥)
𝑓4 (𝑥) = 𝑥𝑓3 ′(𝑥)
…
𝑓𝑞 (𝑥) = 𝑥𝑓𝑞−1 ′(𝑥)
𝑛
(d.i) Show that 𝑓2 (𝑥) = 𝛴 𝑖 2 𝑥 𝑖 .
𝑖=1
[2]
Markscheme

METHOD 1
𝑓2 (𝑥) = 𝑥𝑓1 ′(𝑥)
𝑓1 ′(𝑥) = 12 + 22 𝑥 + (32 𝑥 2 ) + ⋯ + 𝑛 2 𝑥 𝑛−1 (= 1 + 4𝑥 + (9𝑥 2 ) + ⋯ + 𝑛2 𝑥 𝑛−1 )
A1

Note: Award A1 for 𝑥𝑓1 ′(𝑥) = 𝑥(12 + 22 𝑥 + (32 𝑥 2 ) + ⋯ + 𝑛 2 𝑥 𝑛−1 ) (= 𝑥(1 + 4𝑥 + (9𝑥 2 ) +
⋯ + 𝑛 2 𝑥 𝑛−1 ))

𝑥𝑓1 ′(𝑥) = 12 𝑥 + 22 𝑥 2 + (32 𝑥 3 ) + ⋯ + 𝑛 2 𝑥 𝑛 (= 𝑥 + 4𝑥 2 + (9𝑥 3 ) + ⋯ + 𝑛 2 𝑥 𝑛 )

A1
𝑛 𝑛
Note: Award A1 for 𝑓1 ′(𝑥) = 𝛴 𝑖 2𝑥 𝑖−1 and A1 for 𝑥𝑓1 ′(𝑥) = 𝑥 𝛴 𝑖 2𝑥 𝑖−1.
𝑖=1 𝑖=1
The second A1 is dependent on the first A1.
Award a maximum of A0A1 if a general term is not considered.
𝑛
= 𝛴 𝑖 2 𝑥𝑖 AG
𝑖=1

METHOD 2
𝑑
𝑓2 (𝑥) = 𝑥 (𝑥𝑓′(𝑥))
𝑑𝑥
= 𝑥(𝑓′(𝑥) + 𝑥𝑓″(𝑥)) (= 𝑥𝑓′(𝑥) + 𝑥 2 𝑓″(𝑥)) A1
 𝑛 𝑛
= 𝑥 𝛴 𝑖𝑥 𝑖−1 + 𝑥 2 𝛴 𝑖(𝑖 − 1)𝑥 𝑖−2
𝑖=1 𝑖=1
𝑛 𝑛 𝑛
= 𝛴 𝑖𝑥 + 𝛴 𝑖(𝑖 − 1)𝑥 𝑖 (= 𝛴 (𝑖 + 𝑖 2 − 𝑖)𝑥 𝑖 )
𝑖
A1
𝑖=1 𝑖=1 𝑖=1
𝑛
= 𝛴 𝑖 2 𝑥𝑖 AG
𝑖=1

[2 marks]

𝑛
(d.ii) Prove by mathematical induction that 𝑓𝑞 (𝑥) = 𝛴 𝑖 𝑞 𝑥 𝑖 , 𝑞 ∈ ℤ+ .
𝑖=1
[6]
Markscheme

consider 𝑞 = 1
𝑛
𝑓1 (𝑥) = 𝑥 + 2𝑥 2 + ⋯ 𝑛𝑥 𝑛 (reference to part (c)) and 𝑓1 (𝑥) = 𝛴 𝑖𝑥 𝑖 R1
𝑖=1
𝑛
assume true for 𝑞 = 𝑘, (𝑓𝑘 (𝑥) = 𝛴 𝑖 𝑘 𝑥 𝑖 ) M1
𝑖=1

Note: Do not award M1 for statements such as “let 𝑞 = 𝑘” or “𝑞 = 𝑘 is true”. Subsequent marks
after this M1 are independent of this mark and can be awarded.

consider 𝑞 = 𝑘 + 1
𝑓𝑘+1 (𝑥) = 𝑥𝑓𝑘 ′(𝑥) M1
𝑛
= 𝑥 𝛴 𝑖 𝑘+1 𝑥 𝑖−1 OR 𝑥(1 + 2𝑘+1 𝑥 + 3𝑘+1 𝑥 2 + ⋯ + 𝑛 𝑘+1 𝑥 𝑛−1 ) A1
𝑖=1

𝑛 𝑛
Note: Award the above M1 if 𝑓𝑘+1 (𝑥) = 𝑥 𝛴 𝑖 𝑘+1 𝑥 𝑖−1 or 𝑥𝑓𝑘 ′(𝑥) = 𝑥 𝛴 𝑖 𝑘+1 𝑥 𝑖−1 (or
𝑖=1 𝑖=1
equivalent) is stated.
𝑛
= 𝛴 𝑖 𝑘+1 𝑥 𝑖 OR 𝑥 + 2𝑘+1 𝑥 2 + 3𝑘+1 𝑥 3 + ⋯ + 𝑛 𝑘+1 𝑥 𝑛 A1
𝑖=1
since true for 𝑞 = 1 and true for 𝑞 = 𝑘 + 1 if true for 𝑞 = 𝑘, hence true for all 𝑞(∈ ℤ+ )
R1

Note: To obtain the final R1, three of the previous five marks must have been awarded.

[6 marks]

(d.iii) Using sigma notation, write down an expression for 𝑓𝑞 (1).

[1]
Markscheme

𝑓𝑞 (1) = 1𝑞 + 2𝑞 + 3𝑞 + ⋯ + 𝑛 𝑞
𝑛 𝑛
= 𝛴 𝑖 𝑞 (= 𝛴 1𝑖 𝑖 𝑞 ) A1
𝑖=1 𝑖=1

[1 mark]

𝑥 𝑛+1 −1
(e) By considering 𝑓(𝑥) = 1 + 𝑥 + 𝑥 2 + ⋯ + 𝑥 𝑛 as a geometric series, for 𝑥 ≠ 1, show that 𝑓(𝑥) = .
𝑥−1
 [2]
Markscheme
𝑢 (𝑟𝑛 −1)
uses 𝑆𝑛 = 1 𝑟−1 where 𝑟 = 𝑥 and 𝑢1 = 1 M1
clear indication there are (𝑛 + 1) terms R1
𝑥 𝑛+1 −1
𝑓(𝑥) = AG
𝑥−1

[2 marks]

𝑛𝑥 𝑛+2 −(𝑛+1)𝑥 𝑛+1+𝑥

(f) For 𝑥 ≠ 1, show that 𝑓1 (𝑥) = (𝑥−1)2
.
[3]
Markscheme

METHOD 1
𝑓1 (𝑥) = 𝑥𝑓′(𝑥)
(𝑥−1)(𝑛+1)𝑥 𝑛 −1×(𝑥 𝑛+1 −1)
=𝑥 (𝑥−1)2
M1A1

Note: Award M1 for attempting to use the quotient or the product rule to find 𝑓′(𝑥).

(𝑛𝑥+𝑥−𝑛−1)𝑥 𝑛 −(𝑥 𝑛+1 −1) 𝑛𝑥 𝑛+1 −𝑛𝑥 𝑛 −𝑥 𝑛 +1

=𝑥 (𝑥−1)2
(= 𝑥 (𝑥−1)2
) A1
Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

𝑛𝑥 𝑛+2 −(𝑛+1)𝑥 𝑛+1 +𝑥

= (𝑥−1)2
AG

METHOD 2
attempts to form (𝑥 − 1)𝑓1 (𝑥) M1
(𝑥 − 1)𝑓1 (𝑥) = 𝑛𝑥 𝑛+1 − (𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛 )
1 𝑥 𝑛+1 −1
𝑓1 (𝑥) = 𝑥−1 (𝑛𝑥 𝑛+1 − ( − 1)) A1
𝑥−1

1 𝑛𝑥 𝑛+1 (𝑥−1)−𝑥 𝑛+1 +𝑥 1 𝑛𝑥 𝑛+2 −𝑛𝑥 𝑛+1 −𝑥 𝑛+1 +𝑥

𝑓1 (𝑥) = 𝑥−1 ( ) (= 𝑥−1 ( )) A1
𝑥−1 𝑥−1

Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

𝑛𝑥 𝑛+2 −(𝑛+1)𝑥 𝑛+1 +𝑥

𝑓1 (𝑥) = (𝑥−1)2
AG

[3 marks]

(g.i) Show that lim 𝑓1 (𝑥) is in indeterminate form.

𝑥→1
[1]
Markscheme
𝑛−(𝑛+1)+1 0
lim 𝑓1 (𝑥) = (= 0) R1
𝑥→1 0
 Note: Only award R1 for sufficient simplification of the numerator, for example, as shown
above.
Do not award R1 if lim is not referred to or stated.
𝑥→1

[1 mark]

1
(g.ii) Hence, by applying l’Hôpital’s rule, show that lim 𝑓1 (𝑥) = 2 𝑛(𝑛 + 1).
𝑥→1
[5]
Markscheme

attempts to differentiate both the numerator and the denominator M1

𝑛(𝑛+2)𝑥 𝑛+1 −(𝑛+1)2 𝑥 𝑛 +1
lim A1
𝑥→1 2(𝑥−1)

𝑛(𝑛+2)𝑥 𝑛+1 −𝑛(𝑛+1)𝑥 𝑛 −(𝑛+1)𝑥 𝑛+1

Note: Award A1 for ((lim ) ). This form can be used in
𝑥→1 2(𝑥−1)

subsequent work.

(l’Hôpital’s rule applies again since)

𝑛(𝑛+2)𝑥 𝑛+1 −(𝑛+1)2 𝑥 𝑛 +1 0
lim =0 R1
𝑥→1 2(𝑥−1)

Note: Do not award R1 if lim is not referred to or stated. Subsequent marks are independent of
𝑥→1
this R mark.

attempts to differentiate both the numerator and the denominator M1

𝑛(𝑛 + 2)(𝑛 + 1)𝑥 𝑛 − 𝑛(𝑛 + 1)2 𝑥 𝑛−1
lim
𝑥→1 2
𝑛(𝑛 + 2)(𝑛 + 1) − 𝑛(𝑛 + 1)2 𝑛 3 + 3𝑛 2 + 2𝑛 − (𝑛 3 + 2𝑛2 + 𝑛)
= (= )
2 2
𝑛(𝑛+1)((𝑛+2)−(𝑛+1)) 𝑛2 +𝑛
= (= ) A1
2 2
1
= 𝑛(𝑛 + 1) AG
2

[5 marks]

26. 22M.1.SL.TZ1.8
1
Consider the series ln 𝑥 + 𝑝 ln 𝑥 + 3 ln 𝑥 + ⋯, where 𝑥 ∈ ℝ, 𝑥 > 1 and 𝑝 ∈ ℝ, 𝑝 ≠ 0.
Consider the case where the series is geometric.
1
(a.i) Show that 𝑝 = ± .
√3
[2]
Markscheme

EITHER
attempt to use a ratio from consecutive terms M1
1
𝑝 ln 𝑥 ln 𝑥 1 1
3
= 𝑝 ln 𝑥 OR ln 𝑥 = (ln 𝑥)𝑟 2 OR 𝑝 ln 𝑥 = ln 𝑥 (3𝑝)
ln 𝑥 3
 1
Note: Candidates may use ln 𝑥 1 + ln 𝑥 𝑝 + ln 𝑥 3 … and consider the powers of 𝑥 in geometric
sequence
1
𝑝
Award M1 for 1 = 𝑝3 .

OR
1
𝑟 = 𝑝 and 𝑟 2 = 3 M1

THEN
1 1
𝑝2 = 3 OR 𝑟 = ± A1
√3
1
𝑝=± AG
√3

1 1
Note: Award M0A0 for 𝑟 2 = 3 or 𝑝2 = 3 with no other working seen.

[2 marks]

(a.ii) Given that 𝑝 > 0 and 𝑆∞ = 3 + √3, find the value of 𝑥.

[3]
Markscheme
ln 𝑥
1 (= 3 + √3) (A1)
1−
√3
3 √3
ln 𝑥 = 3 − + √3 − OR ln 𝑥 = 3 − √3 + √3 − 1 (⇒ ln 𝑥 = 2) A1
√3 √3
2
𝑥=e A1

[3 marks]

Now consider the case where the series is arithmetic with common difference 𝑑.
2
(b.i) Show that 𝑝 = 3.
[3]
Markscheme

METHOD 1
attempt to find a difference from consecutive terms or from 𝑢2 M1
correct equation A1
1 1
𝑝 ln 𝑥 − ln 𝑥 = 3 ln 𝑥 − 𝑝 ln 𝑥 OR 3 ln 𝑥 = ln 𝑥 + 2(𝑝 ln 𝑥 − ln 𝑥)

1
Note: Candidates may use ln 𝑥 1 + ln 𝑥 𝑝 + ln 𝑥 3 + ⋯ and consider the powers of 𝑥 in arithmetic
sequence.
1
Award M1A1 for 𝑝 − 1 = − 𝑝
3

4 4
2𝑝 ln 𝑥 = 3 ln 𝑥 (⇒ 2𝑝 = 3) A1
2
𝑝=3 AG

METHOD 2
 𝑢1 +𝑢3
attempt to use arithmetic mean 𝑢2 = M1
2
1
ln 𝑥+3ln 𝑥
𝑝 ln 𝑥 = A1
2
4 4
2𝑝 ln 𝑥 = 3 ln 𝑥 (⇒ 2𝑝 = 3) A1
2
𝑝=3 AG

METHOD 3
attempt to find difference using 𝑢3 M1
1 1
ln 𝑥 = ln 𝑥 + 2𝑑 (⇒ 𝑑 = − ln 𝑥)
3 3
1 1 1
𝑢2 = ln 𝑥 + 2 (3 ln 𝑥 − ln 𝑥) OR 𝑝 ln 𝑥 − ln 𝑥 = − 3 ln 𝑥 A1
2
𝑝 ln 𝑥 = 3 ln 𝑥 A1
2
𝑝=3 AG

[3 marks]

(b.ii) Write down 𝑑 in the form 𝑘 ln 𝑥, where 𝑘 ∈ ℚ.

[1]
Markscheme
1
𝑑 = − 3 ln 𝑥 A1

[1 mark]

(b.iii) The sum of the first 𝑛 terms of the series is −3 ln 𝑥.

Find the value of 𝑛.
[6]
Markscheme

METHOD 1
𝑛 1
𝑆𝑛 = [2 ln 𝑥 + [𝑛 − 1] × [− ln 𝑥]]
2 3
attempt to substitute into 𝑆𝑛 and equate to −3 ln 𝑥 (M1)
𝑛 1
[2 ln 𝑥 + [𝑛 − 1] × [− ln 𝑥]] = −3 ln 𝑥
2 3
correct working with 𝑆𝑛 (seen anywhere) (A1)
𝑛 𝑛 1 𝑛(𝑛−1) 𝑛 4−𝑛
[2 ln 𝑥 − ln 𝑥 + ln 𝑥] OR 𝑛 ln 𝑥 − ln 𝑥 OR (ln 𝑥 + ( ) ln 𝑥)
2 3 3 6 2 3
correct equation without ln 𝑥 A1
𝑛 7 𝑛 𝑛(𝑛−1)
( − ) = −3 OR 𝑛 − = −3 or equivalent
2 3 3 6

1 𝑛 7 𝑛
Note: Award as above if the series 1 + 𝑝 + 3 + ⋯ is considered leading to 2 (3 − 3 ) = −3.

attempt to form a quadratic = 0 (M1)

𝑛 2 − 7𝑛 − 18 = 0
 attempt to solve their quadratic (M1)
(𝑛 − 9)(𝑛 + 2) = 0
𝑛=9 A1

METHOD 2
listing the first 7 terms of the sequence (A1)
2 1 1 2
ln 𝑥 + ln 𝑥 + ln 𝑥 + 0 − ln 𝑥 − ln 𝑥 − ln 𝑥 + ⋯
3 3 3 3
recognizing first 7 terms sum to 0 M1
4
8th term is − 3 ln 𝑥 (A1)
5
9th term is − 3 ln 𝑥 (A1)
sum of 8 and 9 term = −3 ln 𝑥
th th (A1)
𝑛=9 A1

[6 marks]

27. 22M.1.SL.TZ2.2
The 𝑛th term of an arithmetic sequence is given by 𝑢𝑛 = 15 − 3𝑛.
(a) State the value of the first term, 𝑢1.
[1]
Markscheme

𝑢1 = 12 A1

[1 mark]

(b) Given that the 𝑛th term of this sequence is −33, find the value of 𝑛.
[2]
Markscheme

15 − 3𝑛 = −33 (A1)
𝑛 = 16 A1

[2 marks]

(c) Find the common difference, 𝑑.

[2]
Markscheme

valid approach to find 𝑑 (M1)

𝑢2 − 𝑢1 = 9 − 12 OR recognize gradient is −3 OR attempts to solve −33 = 12 + 15𝑑
𝑑 = −3 A1

[2 marks]

28. 22M.1.AHL.TZ1.10
1
Consider the series ln 𝑥 + 𝑝 ln 𝑥 + 3 ln 𝑥 + ⋯, where 𝑥 ∈ ℝ, 𝑥 > 1 and 𝑝 ∈ ℝ, 𝑝 ≠ 0.
Consider the case where the series is geometric.
 1
(a.i) Show that 𝑝 = ± .
√3
[2]
Markscheme

EITHER
attempt to use a ratio from consecutive terms M1
1
𝑝 ln 𝑥 ln 𝑥 1 1
3
= 𝑝 ln 𝑥 OR ln 𝑥 = (ln 𝑥)𝑟 2 OR 𝑝 ln 𝑥 = ln 𝑥 (3𝑝)
ln 𝑥 3

1
Note: Candidates may use ln 𝑥 1 + ln 𝑥 𝑝 + ln 𝑥 3 + ⋯ and consider the powers of 𝑥 in geometric
sequence
1
𝑝
Award M1 for 1 = 𝑝3 .

OR
1
𝑟 = 𝑝 and 𝑟 2 = 3 M1

THEN
1 1
𝑝2 = 3 OR 𝑟 = ± A1
√3
1
𝑝=± AG
√3

1 1
Note: Award M0A0 for 𝑟 2 = 3 or 𝑝2 = 3 with no other working seen.

[2 marks]

(a.ii) Hence or otherwise, show that the series is convergent.

[1]
Markscheme

EITHER
1 1
since, |𝑝| = and <1 R1
√3 √3

OR
1
since, |𝑝| = and −1 < 𝑝 < 1 R1
√3

THEN
⇒ the geometric series converges. AG

Note: Accept 𝑟 instead of 𝑝.

Award R0 if both values of 𝑝 not considered.

[1 mark]

(a.iii) Given that 𝑝 > 0 and 𝑆∞ = 3 + √3, find the value of 𝑥.

[3]
Markscheme
 ln 𝑥
1 (= 3 + √3) (A1)
1−
√3
3 √3
ln 𝑥 = 3 − + √3 − OR ln 𝑥 = 3 − √3 + √3 − 1 (⇒ ln 𝑥 = 2) A1
√3 √3
2
𝑥=e A1

[3 marks]

Now consider the case where the series is arithmetic with common difference 𝑑.
2
(b.i) Show that 𝑝 = 3.
[3]
Markscheme

METHOD 1
attempt to find a difference from consecutive terms or from 𝑢2 M1
correct equation A1
1 1
𝑝 ln 𝑥 − ln 𝑥 = 3 ln 𝑥 − 𝑝 ln 𝑥 OR 3 ln 𝑥 = ln 𝑥 + 2(𝑝 ln 𝑥 − ln 𝑥)

1
Note: Candidates may use ln 𝑥 1 + ln 𝑥 𝑝 + ln 𝑥 3 + ⋯ and consider the powers of 𝑥 in arithmetic
sequence.
1
Award M1A1 for 𝑝 − 1 = − 𝑝
3

4 4
2𝑝 ln 𝑥 = 3 ln 𝑥 (⇒ 2𝑝 = 3) A1
2
𝑝=3 AG

METHOD 2
𝑢 +𝑢
attempt to use arithmetic mean 𝑢2 = 1 2 3 M1
1
ln 𝑥+3ln 𝑥
𝑝 ln 𝑥 = A1
2
4 4
2𝑝 ln 𝑥 = 3 ln 𝑥 (⇒ 2𝑝 = 3) A1
2
𝑝=3 AG

METHOD 3
attempt to find difference using 𝑢3 M1
1 1
ln 𝑥 = ln 𝑥 + 2𝑑 (⇒ 𝑑 = − ln 𝑥)
3 3
1 1 1
𝑢2 = ln 𝑥 + 2 (3 ln 𝑥 − ln 𝑥) OR 𝑝 ln 𝑥 − ln 𝑥 = − 3 ln 𝑥 A1
2
𝑝 ln 𝑥 = 3 ln 𝑥 A1
2
𝑝= AG
3

[3 marks]

(b.ii) Write down 𝑑 in the form 𝑘 ln 𝑥, where 𝑘 ∈ ℚ.

[1]
 Markscheme
1
𝑑 = − 3 ln 𝑥 A1

[1 mark]

1
(b.iii) The sum of the first 𝑛 terms of the series is ln (𝑥 3).
Find the value of 𝑛.
[8]
Markscheme

METHOD 1
𝑛 1
𝑆𝑛 = ⌊2 ln 𝑥 + ⌊𝑛 − 1⌋ × ⌊− ln 𝑥⌋⌋
2 3
1
attempt to substitute into 𝑆𝑛 and equate to ln ( 3) (M1)
𝑥
𝑛 1 1
⌊2 ln 𝑥 + ⌊𝑛 − 1⌋ × ⌊− ln 𝑥⌋⌋ = ln ( 3 )
2 3 𝑥
1
ln (𝑥 3) = −ln 𝑥 3 (= ln 𝑥 −3 ) (A1)
= −3 ln 𝑥 (A1)
correct working with 𝑆𝑛 (seen anywhere) (A1)
𝑛 𝑛 1 𝑛(𝑛−1) 𝑛 4−𝑛
⌊2 ln 𝑥 − 3 ln 𝑥 + 3 ln 𝑥⌋ OR 𝑛 ln 𝑥 − 6 ln 𝑥 OR (ln 𝑥 + ( ) ln 𝑥)
2 2 3
correct equation without ln 𝑥 A1
𝑛 7 𝑛 𝑛(𝑛−1)
( − 3 ) = −3 OR 𝑛 − 6 = −3 or equivalent
2 3

1 𝑛 7 𝑛
Note: Award as above if the series 1 + 𝑝 + 3 + ⋯ is considered leading to 2 (3 − 3 ) = −3.

attempt to form a quadratic = 0 (M1)

𝑛 2 − 7𝑛 − 18 = 0
attempt to solve their quadratic (M1)
(𝑛 − 9)(𝑛 + 2) = 0
𝑛=9 A1

METHOD 2
1
ln (𝑥 3) = −ln 𝑥 3 (= ln 𝑥 −3 ) (A1)
= −3 ln 𝑥 (A1)
listing the first 7 terms of the sequence (A1)
2 1 1 2
ln 𝑥 + ln 𝑥 + ln 𝑥 + 0 − ln 𝑥 − ln 𝑥 − ln 𝑥 + ⋯
3 3 3 3
recognizing first 7 terms sum to 0 M1
4
8 term is − ln 𝑥
th (A1)
3
5
9thterm is − ln 𝑥 (A1)
3
sum of 8 and 9 term = −3 ln 𝑥
th th (A1)
𝑛=9 A1

[8 marks]
© International Baccalaureate Organization, 2024