Proof HL Questions [108 marks]

1. [Maximum mark: 5] SPM.1.SL.TZ0.3

(a) Show that (2n − 1)
2
+ (2n + 1)
2
= 8n
2
+ 2, where

n ∈ Z. [2]

Markscheme

attempting to expand the LHS (M1)

LHS = (4n
2
− 4n + 1) + (4n
2
+ 4n + 1) A1

+ 2 (= RHS) AG
2
= 8n

[2 marks]

(b) Hence, or otherwise, prove that the sum of the squares of any
two consecutive odd integers is even. [3]

Markscheme

METHOD 1

recognition that 2n − 1 and 2n + 1 represent two consecutive

odd integers (for n ∈ Z) R1

8n
2
+ 2 = 2 (4n
2
+ 1) A1

valid reason eg divisible by 2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

METHOD 2
 recognition, eg that n and n + 2 represent two consecutive odd integers
(for n ∈ Z) R1

2
n
2
+ (n + 2) = 2 (n
2
+ 2n + 2) A1

valid reason eg divisible by 2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

[3 marks]

2. [Maximum mark: 6] EXN.1.SL.TZ0.4

The first three terms of an arithmetic sequence are u 1 , 5u 1 − 8 and 3u 1 + 8.

(a) Show that u 1 = 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 u 2 − u1 = u3 − u2 (M1)

(5u 1 − 8) − u 1 = (3u 1 + 8) − (5u 1 − 8)

6u 1 = 24 A1

OR

u 1 +u 3
uses u 2 =
2
(M1)
 u 1 +(3u 1 +8)
5u 1 − 8 =
2

3u 1 = 12 A1

THEN

so u 1 = 4 AG

[2 marks]

(b) Prove that the sum of the first n terms of this arithmetic
sequence is a square number. [4]

Markscheme

d = 8 (A1)

uses S n M1
n
= (2u 1 + (n − 1)d)
2

A1
n
Sn = (8 + 8(n − 1))
2

2
= 4n

2
= (2n) 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]

3. [Maximum mark: 8] EXN.1.AHL.TZ0.9

It is given that 2 cos A sin B ≡ sin(A + B) − sin(A − B)

. (Do not prove this identity.)

Using mathematical induction and the above identity, prove that

n
sin 2nθ
Σ cos(2r − 1)θ =
2 sin θ
for n ∈ Z
+
. [8]
r=1

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.

let P(n) be the proposition that

sin 2nθ
Σ cos(2r − 1)θ =
2 sin θ
for
r=1
+
n ∈ Z

considering P(1):

LHS = cos(1)θ = cos θ and

sin 2θ 2 sin θ cos θ
RHS = = = cos θ = LHS
2 sin θ 2 sin θ

so P(1) is true R1

assume P(k) is true, i.e. Σ cos(2r − 1)θ =

sin 2kθ

2 sin θ
(k ∈ Z
+
)
r=1

M1

Note: Award M0 for statements such as “let n = k”.

Note: Subsequent marks after this M1 are independent of this mark and
can be awarded.

considering P(k + 1)

k+1 k

Σ cos(2r − 1)θ = Σ cos(2r − 1)θ + cos(2(k + 1) − 1)θ

r=1 r=1

M1

=
sin 2kθ

2 sin θ
+ cos(2(k + 1) − 1)θ A1

sin 2kθ+2 cos((2k+1)θ) sin θ

=
2 sin θ

sin 2kθ+sin((2k+1)θ+θ)− sin((2k+1)θ−θ)

=
2 sin θ
M1

Note: Award M1 for use of

2 cos A sin B = sin(A + B) − sin(A − B) with

A = (2k + 1)θ and B = θ.

sin 2kθ+sin(2k+2)θ− sin 2kθ

=
2 sin θ
A1

sin 2(k+1)θ
=
2 sin θ
A1

P(k + 1) is true whenever P(k) is true, P(1) is true, so P(n) is true for

n ∈ Z
+
R1

Note: Award the final R1 mark provided at least five of the previous marks
have been awarded.
 [8 marks]

4. [Maximum mark: 6] EXN.2.AHL.TZ0.8

Prove by contradiction that log 2 5 is an irrational number. [6]

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.
p
assume there exist p, q ∈ N where q ≥ 1 such that log 2 5 =
q

M1A1

Note: Award M1 for attempting to write the negation of the statement as

an assumption. Award A1 for a correctly stated assumption.

p
p
log 2 5 =
q
⇒ 5 = 2 q
A1

5
q
= 2
p
A1

EITHER

5 is a factor of 5 q but not a factor of 2 p R1

OR

2 is a factor of 2 p but not a factor of 5 q R1

 OR

5
q
is odd and 2 p is even R1

THEN

no p, q ∈ N (where q ≥ 1) satisfy the equation 5 q = 2

p
and this is a
contradiction R1

so log 2 5 is an irrational number AG

[6 marks]

5. [Maximum mark: 8] EXM.1.SL.TZ0.1

(a) Explain why any integer can be written in the form 4k or
4k + 1 or 4k + 2 or 4k + 3, where k ∈ Z. [2]

Markscheme

Upon division by 4 M1

any integer leaves a remainder of 0, 1, 2 or 3. R1

Hence, any integer can be written in the form 4k or 4k + 1 or 4k + 2 or

4k + 3, where k ∈ Z AG

[2 marks]

(b) Hence prove that the square of any integer can be written in the
form 4t or 4t + 1, where t ∈ Z
+
. [6]
 Markscheme

2
(4k) = 16k
2
= 4t M1A1

2
(4k + 1) = 16k
2
+ 8k + 1 = 4t + 1 M1A1

2
A1
2
(4k + 2) = 16k + 16k + 4 = 4t

2
(4k + 3) = 16k
2
+ 24k + 9 = 4t + 1 A1

Hence, the square of any integer can be written in the form 4t or 4t + 1,

where t ∈ Z
+
. AG

[6 marks]

6. [Maximum mark: 7] 24M.1.AHL.TZ1.7

n!
Using mathematical induction and the definition nC r =
r!(n−r)!
,
n

prove that Σ
r
C
1
=
n+1
C
2
for all n ∈ Z
+
. [7]
r=1

Markscheme

Base case n = 1: LHS =

1
C
1
= 1 and RHS =
2
C
2
= 1, so true for
n = 1 R1

Note: Award R0 if the value of 1C 1 and 2C 2 are not evaluated.

Subsequent marks can still be awarded.

k
r
assume true for n = k ie Σ C
1
=
k+1
C
2
for some k ∈ Z
+
M1
r=1
Note: The assumption of truth must be clear.

Award M0 for statements such as “let n = k” or “ n = k is true”.

Subsequent marks can still be awarded.

consider n = k + 1

k+1

LHS= Σ
r
C
1
r=1

= Σ
r
C
1
+
k+1
C
1
(M1)
r=1

k+1 (k+1)! (k+1)!

= C
2
+
k+1
C
1
OR 2(k−1)!
+
k!
A1

EITHER

attempt to cancel factorials and use a common denominator M1

(k+1)k+2(k+1) (k+2)(k+1)
= (= )
2 2

OR

attempt to use a common denominator M1

k(k+1)! 2(k+1)! (k+2)(k+1)!

= + (= )
2k! 2k! 2k!

THEN

(k+2)! (k+2)!
=
2!k!
(=
2!(k+2−2)!
) A1
 k+2
= C
2

since true for n = 1, and true for n = k implies true for n = k + 1,

therefore true for all n ∈ Z

+
R1

Note: Only award the final R1 if 4 of the previous 6 marks have been
awarded.

[7 marks]

Note: Throughout this question, condone presence of any additional terms

once the first two correct terms are seen.

7. [Maximum mark: 7] 23N.1.AHL.TZ2.6

Prove by mathematical induction that 5 2n
− 2
3n
is divisible by 17
for all n ∈ Z
+
. [7]

Markscheme

base case n = 1 : 5
2
− 2
3
= 25 − 8 = 17 so true for n = 1

A1

assume true for n = k i.e. 5

2k
− 2
3k
= 17s for s ∈ Z OR 5
2k
− 2
3k

is divisible by 17 M1

Note: The assumption of truth must be clear. Do not award the M1 for
statements such as "let n = k" or "n = k is true". Subsequent marks can
still be awarded.

EITHER

consider n = k + 1: M1

2(k+1) 3(k+1)
5 − 2
= (5 )5
2 2k
− (2 )2
3 3k
A1

2k 3k
= (25)5 − (8)2

= (17)5
2k
+ (8)5
2k
− (8)2
3k
OR
(25)5
2k
− (25)2
3k
+ (17)2
3k
A1

= (17)5
2k
+ 8(5
2k
− 2
3k
) OR 25(5 2k − 2
3k
) + (17)2
3k

= (17)5
2k
+ 8(17s) OR 25(17s) + (17)2 3k

= 17(5
2k
+ 8s) OR 17(25s + 2 3k ) which is divisible by 17
A1

OR

(5
2k
− 2
3k
) × 5
2
= 5
2k+2
− 25 × 2
3k
= 17s × 25 M1

= 5
2k+2
− 8 × 2
3k
− 17 × 2
3k
= 17s × 25 A1

2k+2 3k+3 3k
= 5 − 2 − 17 × 2 = 17s × 25

= 5
2(k+1)
− 2
3(k+1)
− 17 × 2
3k
= 17s × 25 A1

2(k+1) 3(k+1) 3k
= 5 − 2 = 17s × 25 + 17 × 2

hence for n = is
2(k+1) 3(k+1) 3k
k + 1 : 5 − 2 = 17(25s + 2 )

divisible by 17 A1

THEN

since true for n = 1, and true for n = k implies true for n = k + 1,

therefore true for all n ∈ 𝕫

+
R1

Note: Only award R1 if 4 of the previous 6 marks have been awarded

 Note: 5 2k and 2 3k may be replaced by 25 k and 8 k throughout.

[7 marks]

8. [Maximum mark: 7] 23M.1.AHL.TZ2.7

n

Use mathematical induction to prove that

r 1
Σ = 1 −
(r+1)! (n+1)!
r=1

for all integers n ≥ 1. [7]

Markscheme

let P(n) be the proposition that Σ

r

(r+1)!
= 1 −
1

(n+1)!
for all
r=1

integers, n ≥ 1

considering P(1):

LHS =
1

2
and RHS =
1

2
and so P(1) is true R1

assume P(k) is true ie, Σ

r

(r+1)!
= 1 −
1

(k+1)!
M1
r=1

Note: Do not award M1 for statements such as “let n = k” or “n = k is

true”. Subsequent marks after this M1 are independent of this mark and can
be awarded.

considering P(k + 1):

k+1 k

Σ
r
= Σ
r
+
k+1
(M1)
(r+1)! (r+1)! ((k+1)+1)!
r=1 r=1

A1
1 k+1
= 1 − +
(k+1)! (K+2)!
 (k+2)−(k+1)
= 1 −
(k+2)!
A1

= 1 −
1

(k+2)!
(= 1 −
1

((k+1+1)!
) A1

P(k + 1) is true whenever P(k) is true and P(1) is true, so P(n) is true

(for all integers, n ≥ 1) R1

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

[7 marks]

9. [Maximum mark: 6] 23M.2.AHL.TZ1.9

Prove by contradiction that p 2 − 8q − 11 ≠ 0, for any p, q ∈ Z. [6]

Markscheme

Assume p 2 − 8q − 11 = 0 , (p, q ∈ Z) M1

Note: This M1 is dependent on the assumption of truth (implied by “assume”

or “suppose that … is true”.)

Subsequent marks should be awarded independently.

EITHER
 = 8q + 11(= 2(4q + 5) + 1) so p odd ⇒ p odd R1
2 2
p

OR

p even ⇒ p − 8q = 11 even which is a contradiction so p is odd

2

R1

Note: This R1 should be awarded for any valid reason to conclude that p
must be odd.

THEN

p = 2k + 1(, k ∈ Z) M1

2
(2k + 1) = 8q + 11

2
4k + 4k + 1 = 8q + 11

4k
2
+ 4k = 8q + 10 (A1)

+ 2k = 4q + 5 or equivalent with one side odd and one side even

2
2k

A1

a contradiction as LHS is even and RHS is odd R1

Note: This R1 is dependent on all previous marks.

Accept correct variations such as work based on p = 2k − 1.

therefore, if p, q ∈ Z then p
2
− 8q − 11 ≠ 0 AG
 [6 marks]

10. [Maximum mark: 5] 22N.1.SL.TZ0.4

Let a be a constant, where a > 1.

(a) 2
a −1
2
2
a +1
2

Show that a 2 + (
2
) = (
2
) . [3]

Markscheme

Note: Award a maximum of M1A0A0 if the candidate manipulates both sides

of the equation (such as moving terms from one side to the other).

METHOD 1 (working with LHS)

2
attempting to expand (a 2 − 1) (do not accept a 4 + 1 or a
4
− 1)

(M1)
4 2 2 4 2
a −2a +1 4a +a −2a +1
LHS = a
2
+
4
or 4
A1
4 2

=
a +2a +1

4
A1

2
2
a +1
= (
2
) (= RHS) AG

Note: Do not award the final A1 if further working contradicts the AG.

METHOD 2 (working with RHS)

2
attempting to expand (a (M1)
2
+ 1)
 4 2
a +2a +1
RHS =
4

2 4 2

=
4a +a −2a +1

4
A1

4 2
a −2a +1
= a
2
+
4
A1

2
2
a −1
= a
2
+ (
2
) (= LHS) AG

Note: Do not award the final A1 if further working contradicts the AG.

[3 marks]

2
a −1
Consider a right-angled triangle with sides of length a, (
2
) and

).
a +1
(
2

(b) Find an expression for the area of the triangle in terms of a. [2]

Markscheme

2
a −1
recognise base and height as a and ( 2
) (may be seen in diagram)
(M1)

correct substitution into triangle area formula A1

2
2 a(a −1) 3

Area =
a a −1
(or equivalent)
a −a
( ) (= = )
2 2 4 4

[2 marks]
11. [Maximum mark: 6] 22M.1.SL.TZ2.3
Consider any three consecutive integers, n − 1, n and n + 1.

(a) Prove that the sum of these three integers is always divisible by
3. [2]

Markscheme

(n − 1) + n + (n + 1) (A1)

= 3n A1

which is always divisible by 3 AG

[2 marks]

(b) Prove that the sum of the squares of these three integers is
never divisible by 3. [4]

Markscheme

2 2 2 2 2 2
(n − 1) + n + (n + 1) (= n − 2n + 1 + n + n + 2n + 1)

A1

2 2
attempts to expand either (n − 1) or (n + 1) (do not accept
− 1 or n + 1) (M1)
2 2
n

= 3n
2
+ 2 A1

2
demonstrating recognition that 2 is not divisible by 3 or 3
seen after
correct expression divided by 3 R1

3n
2
is divisible by 3 and so 3n 2 + 2 is never divisible by 3
 OR the first term is divisible by 3, the second is not
2

OR OR
2 2 3n +2 2 2
3(n + ) = n +
3 3 3

hence the sum of the squares is never divisible by 3 AG

[4 marks]

12. [Maximum mark: 6] 22M.1.AHL.TZ1.8

Consider integers a and b such that a 2 + b
2
is exactly divisible by 4.
Prove by contradiction that a and b cannot both be odd. [6]

Markscheme

Assume that a and b are both odd. M1

Note: Award M0 for statements such as “let a and b be both odd”.

Note: Subsequent marks after this M1 are independent of this mark and can
be awarded.

Then a = 2m + 1 and b = 2n + 1 A1

2 2 2 2
a + b ≡ (2m + 1) + (2n + 1)

= 4m
2
+ 4m + 1 + 4n
2
+ 4n + 1 A1

= 4(m
2
+ m + n
2
+ n) + 2 (A1)

(4(m 2 + m + n 2 + n) is always divisible by 4) but 2 is not divisible

by 4. (or equivalent) R1

⇒ a
2
+ b
2
is not divisible by 4, a contradiction. (or equivalent) R1
 hence a and b cannot both be odd. AG

Note: Award a maximum of M1A0A0(A0)R1R1 for considering identical or

two consecutive odd numbers for a and b.

[6 marks]

13. [Maximum mark: 5] 22M.1.AHL.TZ2.9

Prove by contradiction that the equation 2x 3 + 6x + 1 = 0 has

no integer roots. [5]

Markscheme

METHOD 1 (rearranging the equation)

assume there exists some α ∈ Z such that 2α 3 + 6α + 1 = 0 M1

Note: Award M1 for equivalent statements such as ‘assume that α is an

integer root of 2α 3 + 6α + 1 = 0’. Condone the use of x throughout

the proof.

Award M1 for an assumption involving α 3 + 3α +

1

2
= 0.

Note: Award M0 for statements such as “let’s consider the equation has
integer roots…” ,“let α ∈ Z be a root of 2α 3 + 6α + 1 = 0…”

Note: Subsequent marks after this M1 are independent of this M1 and can
be awarded.

attempts to rearrange their equation into a suitable form M1

EITHER

2α
3
+ 6α = −1 A1

+ 6α is even R1
3
α ∈ Z ⇒ 2α

+ 6α = −1 which is not even and so α cannot be an integer

3
2α

R1

Note: Accept ‘2α 3 + 6α = −1 which gives a contradiction’.

OR

1 = 2(−α
3
− 3α) A1

α ∈ Z ⇒ (−α
3
− 3α) ∈ Z R1

⇒ 1 is even which is not true and so α cannot be an integer R1

Note: Accept ‘⇒ 1 is even which gives a contradiction’.

OR

2
= −α
3
− 3α A1

α ∈ Z ⇒ (−α
3
− 3α) ∈ Z R1

1
− 3α is is not an integer (= ) and so α cannot be an integer
3
−α
2

R1

1
Note: Accept ‘ −α 3 − 3α is not an integer (= 2
) which gives a
contradiction’.
OR

α = −
1
2
2(α +3)
A1

1
α ∈ Z ⇒ − 2
2(α +3)
∈ Z R1

1
− 2
2(α +3)
is not an integer and so α cannot be an integer R1

1
Note: Accept − 2(α 2
+3)
is not an integer which gives a contradiction’.

THEN

so the equation 2x 3 + 6x + 1 = 0 has no integer roots AG

METHOD 2

assume there exists some α ∈ Z such that 2α 3 + 6α + 1 = 0 M1

Note: Award M1 for equivalent statements such as ‘assume that α is an

integer root of 2α 3 + 6α + 1 = 0’. Condone the use of x throughout

the proof. Award M1 for an assumption involving α 3 + 3α +

1

2
= 0

and award subsequent marks based on this.

Note: Award M0 for statements such as “let’s consider the equation has
integer roots…” ,“let α ∈ Z be a root of 2α 3 + 6α + 1 = 0…”

Note: Subsequent marks after this M1 are independent of this M1 and can
be awarded.

let f (x) = 2x
3
+ 6x + 1 (and f (α) = 0)
 + 6 > 0 for all x ∈ R is a (strictly) increasing
2
f ′(x) = 6x ⇒ f

function M1A1

f (0) = 1 and f (−1) = −7 R1

thus f (x) = 0 has only one real root between −1 and 0, which gives a
contradiction

(or therefore, contradicting the assumption that f (α) = 0 for some

α ∈ Z), R1

so the equation 2x 3 + 6x + 1 = 0 has no integer roots AG

[5 marks]

14. [Maximum mark: 7] 21N.1.SL.TZ0.6

(a)
2

Show that 2x − 3 −
6 2x −5x−3
= , x ∈ R, x ≠ 1
x−1 x−1

. [2]

Markscheme

METHOD 1

attempt to write all LHS terms with a common denominator of x − 1

(M1)

6 2x(x−1)−3(x−1)−6 (2x−3)(x−1) 6
2x − 3 −
x−1
=
x−1
OR x−1
−
x−1

2 2

OR A1
2x −2x−3x+3−6 2x −5x+3 6
= −
x−1 x−1 x−1

AG
2x −5x−3
=
x−1
 METHOD 2

attempt to use algebraic division on RHS (M1)

correctly obtains quotient of 2x − 3 and remainder −6 A1

= 2x − 3 −
6

x−1
as required. AG

[2 marks]

(b) Hence or otherwise, solve the equation

6 π
2 sin 2θ − 3 −
sin 2θ−1
= 0 for 0 ≤ θ ≤ π, θ ≠
4
. [5]

Markscheme

2
2 sin 2θ−5 sin 2θ−3
consider the equation sin 2θ−1
= 0 (M1)

2
⇒ 2 sin 2θ − 5 sin 2θ − 3 = 0

EITHER

attempt to factorise in the form (2 sin 2θ + a)(sin 2θ + b)

(M1)

Note: Accept any variable in place of sin 2θ.

(2 sin 2θ + 1)(sin 2θ − 3) = 0

OR

attempt to substitute into quadratic formula (M1)

 5±√ 49
sin 2θ =
4

THEN

1
sin 2θ = −
2
or sin 2θ = 3 (A1)

Note: Award A1 for sin 2θ = −

1

2
only.

7π 11π
one of 6
OR 6
(accept 210 or 330) (A1)

7π 11π
θ =
12
,
12
(must be in radians) A1

Note: Award A0 if additional answers given.

[5 marks]

15. [Maximum mark: 4] 21M.1.SL.TZ2.2

Consider two consecutive positive integers, n and n + 1.

Show that the difference of their squares is equal to the sum of the two
integers. [4]

Markscheme

attempt to subtract squares of integers (M1)

2 2
(n + 1) − n
EITHER

2
correct order of subtraction and correct expansion of (n + 1) , seen
anywhere A1A1

2 2
= n + 2n + 1 − n (= 2n + 1)

OR

correct order of subtraction and correct factorization of difference of

squares A1A1

= (n + 1 − n)(n + 1 + n)(= 2n + 1)

THEN

= n + n + 1 = RHS A1

Note: Do not award final A1 unless all previous working is correct.

which is the sum of n and n + 1 AG

Note: If expansion and order of subtraction are correct, award full marks
for candidates who find the sum of the integers as 2n + 1 and then show
that the difference of the squares (subtracted in the correct order) is
2n + 1.

[4 marks]
16. [Maximum mark: 9] 19M.2.AHL.TZ1.H_8
(a) Solve the inequality x 2 > 2x + 1. [2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.

x < −0.414, x > 2.41 A1A1

(x < 1 − √ 2, x > 1 + √ 2)

Note: Award A1 for −0.414, 2.41 and A1 for correct inequalities.

[2 marks]

(b) Use mathematical induction to prove that 2 n+1 > n

2
for
n ∈ Z, n ⩾ 3. [7]

Markscheme

check for n = 3,

16 > 9 so true when n = 3 A1

assume true for n = k

2
k+1
> k
2
M1

Note: Award M0 for statements such as “let n = k”.

Note: Subsequent marks after this M1 are independent of this mark and can
be awarded.

prove true for n = k + 1

 k+2 k+1
2 = 2 × 2

> 2k
2
M1

= k
2
+ k
2
(M1)

+ 2k + 1 (from part (a)) A1

2
> k

which is true for k ≥ 3 R1

Note: Only award the A1 or the R1 if it is clear why. Alternate methods are
possible.

2
= (K + 1)

hence if true for n = k true for n = k + 1, true for n = 3 so true for all
n≥3 R1

Note: Only award the final R1 provided at least three of the previous marks
are awarded.

[7 marks]

17. [Maximum mark: 6] 18N.1.AHL.TZ0.H_6

Use mathematical induction to prove that
n

∑ r (r!) = (n + 1)! − 1, for n ∈ Z .

+

r=1
[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.

consider n = 1. 1 (1!) = 1 and 2! − 1 = 1 therefore true for n = 1

R1
Note: There must be evidence that n = 1 has been substituted into both
expressions, or an expression such LHS=RHS=1 is used. “therefore true for
n = 1” or an equivalent statement must be seen.

assume true for n = k, (so that ∑ r (r!) = (k + 1)! − 1) M1

r=1

Note: Assumption of truth must be present.

consider n = k + 1

k+1 k

∑ r (r!) = ∑ r (r!) + (k + 1) (k + 1)! (M1)

r=1 r=1

= (k + 1)! − 1 + (k + 1) (k + 1)! A1

= (k + 2) (k + 1)! − 1 M1

Note: M1 is for factorising (k + 1)!

= (k + 2)! − 1

= ((k + 1) + 1)! − 1

so if true for n = k, then also true for n = k + 1, and as true for n = 1

then true for all n (∈ Z

+
) R1

Note: Only award final R1 if all three method marks have been awarded.
Award R0 if the proof is developed from both LHS and RHS.

[6 marks]
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