Revision Paper 5: Binomial Theorem Solutions

Q1 Expand 61x

x . Hence find the term independent of x.

6 2 3 4 5 66 5 4 3 2

6 4 22 4 6

6 6 6 6 61 1 1 1 1 1 11 2 3 4 5

15 6 16 15 20

x x x x x x xx x x x x x x

x x xx x x

Q2 Show that 4(1 2) 17 12 2 .

4 2 3 4(1 2) 1 4 2 6( 2) 4( 2) ( 2)

1 4 2 6(2) 8 2 4

17 12 2

Q3 Find the ratios of the coefficients of the 3rd and 5th terms in the expansion of 411

2x .

22

3

44

5

4 1 3 ,2 2 2

4 1 14 2 16

3 1Ratio of coefficients = 242 16

T x x

T x x

Q4 Find in ascending powers of x, the first three terms in the expansion of 9

2

1 53

xx

.

9 9 8 72

2 2 2 2

18 15 12

91 1 1 15 9 (5 ) (5 ) ...23 3 3 3

1 5 100 ...19683 729 243

x x xx x x x

x x x

Q5 Find the coefficient of x4y4 in the expansion of 8

3yx .

81

44 4 4

5

4 4

83

8 44

8 704 3 81

70coeff of is .81

rr

ryT x

rrr

yT x x y

x y

Q6 Find the first four terms in the expansion of 15(1 )x , in ascending powers of x. Use the result to estimate the value of 1.0215, giving your answer correct to 3 significant figures.

15 2 3

2 3

15 15

2 3

15 15 15(1 ) 1 ...

1 2 3

1 15 105 455 ...

1 1.020.02

1.02 (1 0.02)1 15(0.02) 105(0.02) 455(0.02) ...1.35

x x x x

x x x

xx

Q7 The coefficient of x3 is 448 in the expansion of 8(1 )ax , where a > 0. Hence, find the coefficient of x4.

1

3 3 34

3

4 45

4

8( )

8( ) 56 ,

3

56 44828

( 2 ) 11204

coeff of is 1120.

rrT axr

T ax a x

aa

T x x

x

Q8 Obtain the first four terms in the expansion of 2 6(2 )x x in ascending powers of x.

2 6 6 5 2 4 2 2 3 2 3

2 2 3 4 3

2 2 3 3

2 3

(2 [ ]) 2 6(2 )( ) 15(2 )( ) 20(2 )( ) ...64 6(32)( ) 15(16)( 2 ) 20(8)( ...) ...64 192 192 240 480 160 ...64 192 48 320 ...

x x x x x x x xx x x x x x

x x x x xx x x

Q9 Given that the coefficient of x4 is 3516

in the expansion of 7

(3 ) 12xpx , find the value of p.

73 4

4 4 4

35 35(3 ) 1 (3 ) .... ...2 8 16

35 35 35316 8 16

105 70 3516 16

105 70 351

xpx px x x

x p x x

p

pp

Q10 The first four terms in the expansion of (2 )nqx , where n > 0, are 2 2 3278 1442

x r x sx . Calculate the values of n, r and s.

1 2 2 3 3

2 2 3

1 2

2 2 2 2

3 3

(2 ) 2 2 ( ) 2 ( ) 2 ( ) ...1 2 3

278 1442

2 8 3(2 ) 144 3(2 ) 144 12

27 272 3(2)(144) 82 2 2

2 17283

n n n n n

n

n

n

n

n n nqx qx qx qx

x r x sx

nn q q q

nq r r r

nq s s

Q11 1 x2

n

1 nx2 n(n1)

8x2 ...

1 nx

2 n(n1)

8x2 ... 1 ax 7x2 ...

Equating coefficients of x2:

2

( 1) 78

56 0( 8)( 7) 0

n n

n nn n

n = 8 or n = 7 (N.A.)

Hence n = 8.

Q12 For8

12x , the general term is given by 1

82

r

rxT

r .

For the term in x3, r = 3.

T4 83

x2

3

7x3 Hence the coefficient of x3 is 7.

Q13

2 3 5 25 51

2

4 3 52 23 3 2 53 22 3 3 5

4

2 3 4 3 5

3280 3 240 120 3 90 9 3

362 209 3

52 3 362 209 3

5 5

22

2 3 2 3 362 209 3 362 209 3

362 209 3

1

Hence 5

5

12 3 . (shown)2 3

Q14 (1 x)6 1 61

(x)

62

(x)

2 63

(x)

3...

= 1 6x + 15x2 20x3 +

(1 x 2x2 )6 1 2x2 x 6

Replace x by 2x2 x.

1 6(2x2 x) + 15(2x2 x)2 20(2x2 x)3 +

= 1 12x2 + 6x + 15(4x4 4x3 + x2) 20( x3 + )

+

= 1 12x2 + 6x 60x3 + 15x2 + 20x3 +

= 1 + 6x + 3x2 40x3 +

Q15

(1 x)5(1 4x)4

1 5x 10x2 ... 1 4(4x) 6(4x)2 ... 1 5x 10x2 ... 116x 96x2 ... = 1 16x + 96x2 + 5x - 80x2 + 10x2 +

= 1 11x+ 26x2 +

Q16

6 8 6 22

2

62 2

2 2

8

1 1 1 1

1 12

Coefficient of 2 6 12

a a a aa a a a

a aa a

a

Q17

5 2

2 2

2 2

2

2

5(1 ) 1 (1 ) 1 5 ...

2

(1 ) 1 5 10 ...

1 (5 ) (10 +5 ) ...5 6 (1)

10 +5 0 (2)6 5 (3)

10 5( 6 5 ) 0( 2) 0

0 (rej.) or 216

px qx px qx qx

px qx q x

q p x q pq xq p

q pqp q

q q qq q

q qp

Q18

22 2

( 1)1 1 1 1 ...2 2 4

( 1)12 4 2

nx nx n n xkx kx

nx n n x knxkx

0 (1)2

( 1) 15 (2)4 2 4

nk

n n kn

Sub (1) into (2):

2( 1) 154 4 4

15

n n n

n

152

k

Q19

21

2 0

2

2 1

2Constant term 1

2 !1

! !2 !

1!

rn r

r

n r r

n

n

n

nT x

r x

x xn r

nnn

n nn

n

Q20

1 2 2 3 3

1 2 2 3 3

1 2 2 3 2

1 2 2

...2 3

...2 3

...2 3

is divisible by

n n n n n n

n n n n n n

n n n n n

n n

n na b a na b a b a b b

n na b a na b a b a b b

n na b a a nb b a a b b

a b a a nb b M b

10 14

10 9 5

10 9

10 10 1

2

5 2

5 2 2

5 2 32

2 3 2 2 10(3)

39 is divisble by 9

MM