MA1301 Introductory Mathematics

Tutorial 2 Solution

TAN BAN PIN

National University of Singapore

TAN BAN PIN MA1301 Introductory Mathematics 1 / 29

Question 1

(a) Find the smallest positive integer n for which the sum of the first n
terms of the following arithmetic sequence

5, 8, 11, 14, · · ·

exceeds 1500.
(b) Find the smallest positive integer n for which the sum of the first n
terms of the geometric sequence

64, 32, 16, · · ·

exceeds 127.95.

TAN BAN PIN MA1301 Introductory Mathematics 2 / 29

Question 1(a) Suggested Solution
Note that the terms form an arithmetic sequence with first term 5 and
common difference 3. Then the sum of the first n terms is
n n
[2 × 5 + (n − 1) × 3] = (3n + 7).
2 2
Then
n
(3n + 7) > 1500 ⇒ 3n2 + 7n − 3000 > 0.
2
The two roots of 3n2 + 7n − 3000 = 0 are
p √
−7 + 72 − 4 · 3 · (−3000) −7 + 36049
= ≈ −32.8,
p 2·3 6
√
−7 − 72 − 4 · 3 · (−3000) −7 − 36049
= ≈ 30.5.
2·3 6
Then
3n2 + 7n − 3000 > 0 ⇒ n < −32.8 or n > 30.5.
Since n is a natural number, n > 30.5 and thus n > 31.
TAN BAN PIN MA1301 Introductory Mathematics 3 / 29
Question 1(b) Suggested Solution

Note that the terms form a geometric sequence with first term 64 and
common ratio 0.5. Then the sum of the first n terms is
1 − 0.5n
64 × = 128(1 − 0.5n ).
1 − 0.5
Then

128(1 − 0.5n ) > 127.95 ⇒ 0.5n < 0.000390625 ⇒ 2n > 2560

ln 2560
⇒n> ≈ 11.3 ⇒ n > 12.
ln 2

TAN BAN PIN MA1301 Introductory Mathematics 4 / 29

Question 2

Evaluate the following telescoping sums.

99  
X r+1
(a) lg .
r
r=1

N
X 1
(b) .
4r2 − 1
r=2

TAN BAN PIN MA1301 Introductory Mathematics 5 / 29

Question 2(a) Suggested Solution
Note that lg X
Y = lg X − lg Y .

Thus, lg r+1


r = lg(r + 1) − lg r = − lg r + lg(r + 1).

Hence,
99   99
X r+1 X
lg = (− lg r + lg(r + 1))
r
r=1 r=1
= (− lg 1 + lg 2) + (− lg 2 + lg 3) + (− lg 3 + lg 4) + · · ·
+ (− lg 98 + lg 99) + (− lg 99 + lg 100)
= − lg 1 + (lg 2 − lg 2) + (lg 3 − lg 3) + · · ·
+ (lg 99 − lg 99) + lg 100
= − lg 1 + lg 100
=0+2
= 2.
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Question 2(b) Suggested Solution

Note that 4r2 − 1 = (2r − 1)(2r + 1). Write

1 A B (2A + 2B)r + (A − B)
= + = .
4r2 −1 2r − 1 2r + 1 4r2 − 1
Compare the coefficients of the numerator:

2A + 2B = 0 and A − B = 1.

Substitute A = B + 1 into 2A + 2B = 0 to get

1 1 1
0 = 2(B + 1) + 2B = 4B + 2 ⇒ B = − ⇒ A = − + 1 = .
2 2 2
Therefore,
1 1 1
= − .
4r2 −1 2(2r − 1) 2(2r + 1)

TAN BAN PIN MA1301 Introductory Mathematics 7 / 29

Question 2(b) Suggested Solution
N N  
X 1 X 1 1
= −
4r2 − 1 2(2r − 1) 2(2r + 1)
r=2 r=2
     
1 1 1 1 1 1
= − + − + − + ···
2·3 2·5 2·5 2·7 2·7 2·9
 
1 1
+ −
2 · (2N − 3) 2 · (2N − 1)
 
1 1
+ −
2 · (2N − 1) 2 · (2N + 1)
   
1 1 1 1 1
= + − + + − + + ···
2·3 2·5 2·5 2·7 2·7
 
1 1 1
+ − + −
2 · (2N − 1) 2 · (2N − 1) 2(2N + 1)
1 1 1 1
= − = − .
2 · 3 2(2N + 1) 6 2(2N + 1)
TAN BAN PIN MA1301 Introductory Mathematics 8 / 29
Question 3

(i) Show that for any positive integer r,

(r + 1)! − r! = r(r!).

(ii) Use (i) to show that

m
X
r(r!) = (m + 1)! − 1.
r=1

TAN BAN PIN MA1301 Introductory Mathematics 9 / 29

Question 3(i) Suggested Solution

Note that for any positive integer r,

(r + 1)! = 1 · 2 · 3 · · · (r − 1) · r · (r + 1) = r!(r + 1).

Then

(r + 1)! − r! = (r + 1)(r!) − r! = (r + 1 − 1)(r!) = r(r!).

TAN BAN PIN MA1301 Introductory Mathematics 10 / 29

Question 3(ii) Suggested Solution

Since r(r!) = (r + 1)! − r! for all positive integer r,

m
X m
X
r(r!) = [−r! + (r + 1)!]
r=1 r=1
= (−1! + 2!) + (−2! + 3!) + (−3! + 4!) + · · ·
+ [−(m − 1)! + m!] + [−m! + (m + 1)!]
= −1! + (2! − 2!) + (3! − 3!) + · · · + (m! − m!) + (m + 1)!
= −1! + (m + 1)! = (m + 1)! − 1.

TAN BAN PIN MA1301 Introductory Mathematics 11 / 29

Question 4

(i) Obtain the binomial expansion of (2 − 3x)6 in ascending powers of x.

(ii) Determine the coefficient of x3 in the expansion of (2 − 3x)8 .

(iii) Find the value of the constant a for which the coefficient of x in the
expansion of (1 + ax)(2 − 3x)6 is zero.

TAN BAN PIN MA1301 Introductory Mathematics 12 / 29

Question 4(i) Suggested Solution

By binomial theorem,

(2 − 3x)6 = [2 + (−3x)]6
     
6 6 0 6 5 1 6 4
= 2 (−3x) + 2 (−3x) + 2 (−3x)2
0 1 2
     
6 3 6 2 6 1
+ 2 (−3x)3 + 2 (−3x)4 + 2 (−3x)5
3 4 5
 
6 0
+ 2 (−3x)6
6
= 1 · 64 · 1 + 6 · 32 · (−3x) + 15 · 16 · 9x2 + 20 · 8 · (−27x3 )
+ 15 · 4 · 81x4 + 6 · 2 · (−243x5 ) + 1 · 1 · 729x6
= 64 − 576x + 2160x2 − 4320x3 + 4860x4 − 2916x5 + 729x6 .

TAN BAN PIN MA1301 Introductory Mathematics 13 / 29

Question 4(ii) Suggested Solution

A term in the expansion of (2 − 3x)8 = [2 + (−3x)]8 is given by

   
8 8−r 8 8−r
2 (−3x)r = 2 (−3)r xr
r r

So the coefficient of x3 is
 
8
· 28−3 · (−3)3 = 56 · 32 · (−27) = −48384.
3

TAN BAN PIN MA1301 Introductory Mathematics 14 / 29

Question 4(iii) Suggested Solution

Note that x = x · 1 = 1 · x. Then

coef. of x in (1 + ax)(2 − 3x)6 = (coef. of x in (1 + ax))

× (coef. of 1 in (2 − 3x)6 )
+ (coef. of 1 in (1 + ax))
× (coef. of x in (2 − 3x)6 )
= a · 64 + 1 · (−576) = 64a − 576.

Then 64a − 576 = 0 implies a = 9.

TAN BAN PIN MA1301 Introductory Mathematics 15 / 29

Question 5

Find the constant term in the binomial expansion of the following.

12
(a) 3x − x1 .
 9
x √4
(b) 2 + x
.

TAN BAN PIN MA1301 Introductory Mathematics 16 / 29

Question 5(a) Suggested Solution

Note that the constant term is the coefficient of x0 .

12
A term of 3x − x1 is given by

1 r
     
12 12−r 12 12−r
(3x) − = 3 (−1)r · x(12−r)−r .
r x r

Let (12 − r) − r = 0. Then r = 6. So the constant term is

 
12 12−6
3 (−1)6 = 924 · 729 · 1 = 673596.
6

TAN BAN PIN MA1301 Introductory Mathematics 17 / 29

Question 5(b) Suggested Solution

 9
x √4
A term of 2 + x
is given by

4 r
       r
9 x 9−r 9 4 r
√ = 9−r
· x(9−r)− 2 .
r 2 x r 2
r
Let (9 − r) − 2 = 0. Then r = 6. So the constant term is
  6
9 4 4096
9−6
= 84 · = 43008.
6 2 8

TAN BAN PIN MA1301 Introductory Mathematics 18 / 29

Question 6

Use the binomial theorem to show that for all natural number n,
       
n n n n
+ + + ··· + = 2n .
0 1 2 n

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Question 6 Suggested Solution

Recall the binomial theorem:

         
n n n n n−1 n n−2 2 n n−1 n n
(a+b) = a + a b+ a b +· · ·+ ab + b .
0 1 2 n−1 n

Let a = b = 1. Then
         
n n n n−1 n n−2 2 n n n
(1+1)n = 1 + 1 1+ 1 1 +· · ·+ 1·1n−1 + 1 .
0 1 2 n−1 n

That is,        
n n n n
+ + + ··· + = 2n .
0 1 2 n

TAN BAN PIN MA1301 Introductory Mathematics 20 / 29

Question 7

√
(i) Expand 1 − x up to and including the term in x2 .
1
√ 10837
(ii) By taking x = 64 in the expansion of (i), deduce that 7≈ 4096 .

TAN BAN PIN MA1301 Introductory Mathematics 21 / 29

Question 7(i) Suggested Solution

Recall that for −1 < x < 1,

n(n − 1) 2
(1 + x)n = 1 + nx + x + ··· .
2!
√ 1
Note that 1 − x = [1 + (−x)] 2 . Then

√ 1 1
1 ( − 1) 1 1
1 − x = 1 + (−x) + 2 2 (−x)2 + · · · = 1 − x − x2 + · · · .
2 2! 2 8

TAN BAN PIN MA1301 Introductory Mathematics 22 / 29

Question 7(ii) Suggested Solution

1
Let x = 64 . Then
r  2
1 1 1 1 1 32511
1− ≈1− · − · = .
64 2 64 8 64 32768
q
1
q q
32 ·7
√
Note that 1 − 64 = 63 64 = 82
= 38 7. Then

√ 32511 8 10837
7≈ · = .
32768 3 4096

TAN BAN PIN MA1301 Introductory Mathematics 23 / 29

Question 8

x
(i) Express x2 −3x+2
in the partial fraction form.

(ii) Show that, if x is so small that x3 and higher powers of x can be

neglected, then
x 1 3
2
≈ x + x2 .
x − 3x + 2 2 4

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Question 8(i) Suggested Solution

Note that x2 − 3x + 2 = (x − 1)(x − 2). Suppose

x A B (A + B)x + (−2A − B)
= + = .
x2 − 3x + 2 x−1 x−2 x2 − 3x + 2
Compare the coefficients of the numerator:

A + B = 1 and − 2A − B = 0.

Add these two equations:

(A + B) + (−2A − B) = 1 + 0.

Then −A = 1 ⇒ A = −1 ⇒ B = 2, and thus

x 1 2
=− + .
x2 − 3x + 2 x−1 x−2

TAN BAN PIN MA1301 Introductory Mathematics 25 / 29

Question 8(ii) Suggested Solution
1
Since − x−1 = [1 + (−x)]−1 ,

1 (−1)(−1 − 1)
− ≈ 1 + (−1)(−x) + (−x)2 = 1 + x + x2 .
x−1 2!
2 1

−1
= − 1−1 x = − 1 + − x2

Since x−2 = x −1 ,
2 2
  x  (−1)(−1 − 1)  x 2 
2 1 1
≈ − 1 + (−1) − + − = −1 − x − x2 .
x−2 2 2! 2 2 4
Therefore, for x small enough,
x 1 2
=− +
x2 − 3x + 2 x − 1 x − 2 
1 1
≈ (1 + x + x ) + −1 − x − x2
2
2 4
1 3
= x + x2 .
2 4
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Question 9

Show that for sufficiently small x,

r
4−x 5 55
≈ 2 − x + x2 .
1+x 4 64

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Question 9 Suggested Solution

√ x

 1
= 2 1 + − x4 2 ,
p 
Since 4−x=2 1− 4
" #
√ 1  x  12 ( 12 − 1)  x 2 1 1
4−x≈2 1+ − + − = 2 − x − x2 .
2 4 2! 4 4 64

1
Since √1
1+x
= (1 + x)− 2 ,

(− 12 )(− 21 − 1) 2
 
1 1 1 3
√ ≈1+ − x+ x = 1 − x + x2 .
1+x 2 2! 2 8

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Question 9 Suggested Solution

q
4−x
√
Note that = 4−x· √1 . Then for sufficiently small x,
1+x 1+x
r   
4−x 1 1 2 1 3 2
≈ 2− x− x 1− x+ x
1+x 4 64 2 8
     
1 1
=2·1+ 2 − + − 1 x
2 4
      
3 1 1 1
+ 2· + − − + − 1 x2 + · · ·
8 4 2 64
5 55
≈ 2 − x + x2 .
4 64

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