64
Chapter 4 The Binomial Theorem
CHAPTER 4 Exercise 4A (p.84) 1. 3!= 3 × 2 × 1 =6
11. (n − 1)! − (n − 2)! = (n − 1) × (n − 2)! − (n − 2)! = (n − 2)!(n − 1 − 1) = (n − 2)[(n − 2)!] n +1 C3
=
13.
n +1 Cn
=
2. 6!= 6 × 5 × 4 × 3 × 2 × 1 = 720 3.
4.
7! 7 × 6 × 5 × 4! = 4! 4! = 7×6×5 = 210 10! 10 × 9 × 8! = 8! 8! = 90
5. 7! − 6! = 7 × 6! − 6! = 6! (7 − 1) = 6! × 6 = 6 × 6 × 5 × 4 × 3 × 2 ×1 = 4 320 6. 4! + 5! = 4! + 5 × 4! = 4! (1 + 5) = 6 × 4 × 3 × 2 ×1 = 144 7.
8.
9.
10.
5 C3
6 C5
5! 3!(5 − 3)! 5! = 3! 2! 5 × 4 × 3! = 3! 2! = 10 =
6! 5!(6 − 5)! 6 × 5! = 5!1! =6 =
12! 9!(12 − 9)! 12 × 11 × 10 × 9! = 9! 3! = 220
12 C9 =
(n + 1)! (n + 1) × n × (n − 1) × (n − 2)! = (n − 2)! (n − 2)! = n(n + 1)(n − 1)
(n + 1)! 3!(n + 1 − 3)! (n + 1)! = 3!(n − 2)! (n + 1)n(n − 1)(n − 2)! = 3!(n − 2)! 1 = n(n + 1)(n − 1) 6
12.
(n + 1)! n!(n + 1 − n)! (n + 1) × n! = n!1! = n +1
n! (n − 2)![n − (n − 2)!] n! = (n − 2)! 2! n(n − 1)(n − 2)! = 2(n − 2)! 1 = n(n − 1) 2
14. n Cn − 2 =
Exercise 4B (p.90) 1. ( a + b)3 = ( a)3 + 3C1 ( a)2 (b) + 3C2 ( a)(b)2 + b 3 = a 3 + 3a 2 b + 3ab 2 + b 3
2. ( x + 2 y)5
= ( x )5 + 5C1 ( x ) 4 (2 y) + 5C2 ( x )3 (2 y)2 + 5C3 ( x )2 (2 y)3 + 5C4 ( x )(2 y) 4 + (2 y)5 = ( x )5 + 5( x ) 4 (2 y) + 10( x )3 (2 y)2 + 10( x )2 (2 y)3 + 5( x )(2 y) 4 + (2 y)5 = x 5 + 10 x 4 y + 40 x 3 y 2 + 80 x 2 y 3 + 80 xy 4 + 32 y 5 3. (2 x − 3) 4 = (2 x ) 4 + 4 C1 (2 x )3 ( −3) + 4 C2 (2 x )2 ( −3)2 + 4 C3 (2 x )( −3)3 + ( −3) 4 = 16 x 4 − 96 x 3 + 216 x 2 − 216 x + 81
Chapter 4 The Binomial Theorem
65
9. (2 x + y)10
4. (3 x − 2 y)6 = (3 x )6 + 6 C1 (3 x )5 ( −2 y) + 6 C2 (3 x ) 4 ( −2 y)2
The general term in the expansion
+ 6 C3 (3 x )3 ( −2 y)3 + 6 C4 (3 x )2 ( −2 y) 4
= 10 Cr (2 x )10 − r ( y)r
+ 6 C5 (3 x )( −2 y)5 + ( −2 y)6
∴ The term in x 3 y 7 = 10 C7 (2 x )3 y 7
= 729 x 6 − 2 916 x 5 y + 4 860 x 4 y 2 − 4 320 x 3 y 3 + 2 160 x y − 576 xy + 64 y 2 4
5
6
1 5. ( x − )5 x 1 1 = ( x )5 + 5C1 ( x ) 4 ( − ) + 5C2 ( x )3 ( − )2 x x 1 3 1 4 1 2 + 5C3 ( x ) ( − ) + 5C4 ( x )( − ) + ( − )5 x x x 5 4 1 3 1 2 2 1 3 = x − 5 x ( ) + 10 x ( ) − 10 x ( ) x x x 1 4 1 + 5x(− ) − 5 x x 5 = x − 5 x 3 + 10 x − 10 x −1 + 5 x −3 − x −5
∴ The coefficient of x 3 y 7 = 10 C7 ⋅ 2 3 = 960 1 10. ( − 3 x )8 2 The general term in the expansion 1 = 8Cr ( )8 − r ( −3 x )r 2 1 ∴ The term in x 3 = 8C3 ( )5 ( −3 x )3 2 1 ∴ The coefficient of x 3 = 8C3 ( )5 ( −3)3 2 189 =− 4 11. The general term in the expansion
x 2 6. ( + )6 2 x x x 2 x 2 = ( )6 + 6 C1 ( )5 ( ) + 6 C2 ( ) 4 ( )2 2 2 x 2 x x 3 2 3 x 2 2 4 + 6 C3 ( ) ( ) + 6 C4 ( ) ( ) 2 x 2 x x 2 5 2 6 + 6 C5 ( )( ) + ( ) x 2 x x 2 x 6 x 5 2 x 2 = ( ) + 6( ) ( ) + 15( ) 4 ( )2 + 20( )3 ( )3 2 2 x 2 x 2 x x 2 2 4 x 2 5 2 6 + 15( ) ( ) + 6( )( ) + ( ) 2 x x 2 x 1 6 3 4 15 2 60 96 64 = x + x + x + 20 + 2 + 4 + 6 64 8 4 x x x
7. ( x + 2)8 The general term in the expansion = 8Cr ( x )8 − r (2)r ∴ The term in x 4 = 8C4 x 4 (2) 4 ∴ The coefficient of x 4 = 8C4 2 4 = 1 120 8. (3 x − 2) 7 The general term in the expansion = 7Cr (3 x ) 7 − r ( −2)r ∴ The term in x 5 = 7C2 (3 x )5 ( −2)2 ∴ The coefficient of x 5 = 7C2 ⋅ 35 ( −2)2 = 20 412
(2 − x )6 = 6 Cr (2)6 − r ( − x )r ∴ The term in x 4 = 6 C4 (2)2 ( − x ) 4 ∴ The coefficient of x 4 = 6 C4 (2)2 ( −1) 4 = 60 The general term in the expansion (2 x − 1)9 = 9 Cr (2 x )9 − r ( −1)r ∴ The term in x 4 = 9 C5 (2 x ) 4 ( −1)5 ∴ The coefficient of x 4 = − 2 016 ∴ The coefficient of x 4 in the expansion of (2 − x )6 − (2 x − 1)9 = 60 + 2 016 = 2 076
12. The general term in the expansion 1 1 ( − 6 x )6 = 6 Cr ( )6 − r ( −6 x )r 3x 3x 1 = 6 Cr ( )6 − r ( −6)r x 2 r − 6 3 It is the constant term when 2 r − 6 = 0 , i.e r = 3 . 1 ∴ The constant term = 6 C3 ( )3 ( −6)3 = −160 3 13. The general term in the expansion 1 1 ( 4 x 2 − )9 = 9 Cr ( 4 x 2 )9 − r ( − )r 2x 2x 1 = 9 Cr ( 4)9 − r ( − )r x18 − 3r 2 It is the constant term when 18 − 3r = 0 , i.e r = 6 . 1 ∴ The constant term = 9 C6 ( 4)3 ( − )6 = 84 2
66
Chapter 4 The Binomial Theorem
14. The general term in the expansion (1 + x )
24
=
24 Cr x
r
∴ The coefficient of x i.e. Br = 24 Cr Br + 2 = Br
24 Cr + 2 24 Cr
24! ( r + 2 )![ 24 − ( r + 2 )!] 24! r!( 24 − r )!
=
r
24 Cr
=
57 7
=
57 7
57 (24 − r )! = 7 (r + 2)(r + 1)(22 − r )! (24 − r )(23 − r ) 57 = 7 (r + 2)(r + 1)
The coefficient of the third term = 4 ⋅ n C2 4 ⋅ n C2 = 60 n! = 15 2!(n − 2)! n(n − 1) = 30 n 2 − n − 30 = 0 (n − 6)(n + 5) = 0 n = 6 or −5 (rejected)
The general term in the expansion (1 + 2 x 2 )6 = 6 Cr (2 x 2 )r ∴ The term in x 8 = 6 C4 (2 x 2 ) 4 ∴ The coefficient of x 8 = 240
57(r 2 + 3r + 2) = 7(552 − 47r + r 2 ) 50 r 2 + 500 r − 3 750 = 0 r + 10 r − 75 = 0 (r + 15)(r − 5) = 0 2
Q ∴
r > 0 , r = −15 (rejected) r=5
1 15. (1 − 2 x )9 (1 + )3 x = [1 + 9( −2 x ) + 36( −2 x )2 + 84( −2 x )3 + L] 1 1 1 [1 + 3( ) + 3( )2 + ( )3 ] x x x 2 = (1 − 18 x + 144 x − 672 x 3 + L) 3 3 1 (1 + + 2 + 3 ) x x x ∴ The constant term = 1 + ( −18)(3) + 144(3) + ( −672) = −293
18. (1 + mx 2 ) n = 1 + n C1 ( mx 2 ) + n C2 ( mx 2 )2 + L Comparing coefficients of x 2 and x 4 respectively,
m ⋅ n C1 = 14 mn = 14 .............................(1) m 2 ⋅ n C2 = 21m 2 n(n − 1) m2 ⋅ = 21m 2 2 n 2 − n − 42 = 0 (n + 6)(n − 7) = 0 n = 7 or Put n = 7 into (1), 7m = 14 m=2 ∴ m= 2
−6 (rejected)
19. The general term in the expansion (1 + x ) n = n Cr x r ∴ The coefficient of x 4 =
n C4
= [1 + 3( −5 x ) + 3( −5 x )2 + L]
∴ The coefficient of x 5 =
n C5
[1 + 6(2 x ) + 15(2 x )2 + L]
∴ The coefficient of x 6 =
n C6
16. (1 − 5 x )3 (1 + 2 x )6
= (1 − 15 x + 75 x + L) 2
(1 + 12 x + 60 x 2 + L) = 1 + ( −15 + 12) x + (75 − 15 × 12 + 60) x 2 + L = 1 − 3 x − 45 x 2 + L
∴
a = −3 , b = −45
17. (2 x + 1) = (1 + 2 x ) 2
n
2 n
= 1 + n C1 (2 x 2 ) + n C2 (2 x 2 )2 + L
n C4 + n C6 = 2 n C5 n! n! 2(n!) + = 4!(n − 4)! 6!(n − 6)! 5!(n − 5)! 30 (n − 4)(n − 5) 12(n − 4) + = (n − 4)! (n − 4)! (n − 4)! 30 + (n − 4)(n − 5) = 12(n − 4)
n 2 − 21n + 98 = 0 (n − 14)(n − 7) = 0 n = 7 or 14
Chapter 4 The Binomial Theorem
4. (3 − x + x 2 )8
20. (1 + px )(1 + qx )5
= [3 − x (1 − x )]8
= (1 + px )[1 + 5(qx ) + 10(qx )2 + L]
= 38 − 8 ⋅ 37 x (1 − x ) + 28 ⋅ 36 x 2 (1 − x )2
= (1 + px )(1 + 5qx + 10 q 2 x 2 + L)
−56 ⋅ 35 x 3 (1 − x )3 + L
Comparing the coefficients of x,
= 6 561 − 17 496 x + 17 496 x 2
5q + p = −6 ...........(1)
+20 412 x 2 (1 − 2 x + L) − 13 608 x 3 (1 + L)
Comparing the coefficients of x 2 ,
= 6 561 − 17 496 x + 37 908 x 2 − 54 432 x 3 + L
10 q + 5 pq = 0 ......(2) 2
Put p = −6 − 5q into (2),
5. (1 − 4 x + x 2 )9
10 q 2 + 5( −6 − 5q )q = 0
= [1 − x ( 4 − x )]9
10 q 2 − 30 q − 25q 2 = 0
= 1 − 9 x ( 4 − x ) + 36 x 2 ( 4 − x )2 − L = 1 − 36 x + 9 x 2 + 36 x 2 (16 + L) + L
15q 2 + 30 q = 0 q( q + 2 ) = 0 q = −2 or 0 (rejected)
= 1 − 36 x + 9 x 2 + 576 x 2 + L = 1 − 36 x + 585 x 2 + L
Put q = −2 into p = −6 − 5q ,
Coefficient of x 2 = 585
p = −6 − 5( −2) =4
6. (2 − x + x 2 )10
21 − 22. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
= [2 − x (1 − x )]10 = 210 − 10 ⋅ 2 9 x (1 − x ) + 45 ⋅ 2 8 x 2 (1 − x )2 − 120 ⋅ 2 7 x 3 (1 − x )3 + L = 1 024 − 5 120 x (1 − x ) + 11 520 x 2
Exercise 4C (p.93)
(1 − 2 x + L) − 15 360 x 3 (1 + L) + L
1. (1 + x + 3 x 2 )3
= 1 024 − 5 120 x + 5 120 x 2 + 11 520 x 2
= [1 + x (1 + 3 x )]
3
= 1 + 3 x (1 + 3 x ) + 3 x (1 + 3 x ) + x (1 + 3 x ) 2
2
3
= 1 + 3 x + 9 x 2 + 3 x 2 (1 + 6 x + L) + x 3 (1 + L) = 1 + 3 x + 12 x + 19 x + L 2
3
− 23 040 x 3 − 15 360 x 3 + L 3
= 1 024 − 5 120 x + 16 640 x 2 − 38 400 x 3 + L Coefficient of x 3 = −38 400 1 + 6 x )3 3x 1 1 = [1 − ( − 6 x 2 )]3 x 3 1 1 1 1 = 1 − 3 ⋅ ( − 6 x 2 ) + 3 ⋅ 2 ( − 6 x 2 )2 x 3 x 3 1 1 2 3 − 3 ( − 6x ) x 3 3 1 3 1 = 1 − ( − 6 x 2 ) + 2 ( − 4 x 2 + 36 x 4 ) + L x 3 x 9
7. (1 −
2. (1 + x − 2 x 2 )6
= [1 + x (1 − 2 x )]6 = 1 + 6 x (1 − 2 x ) + 15 x 2 (1 − 2 x )2 +20 x 3 (1 − 2 x )3 + L = 1 + 6 x − 12 x 2 + 15 x 2 (1 − 4 x + L) +20 x 3 (1 + L) = 1 + 6 x + 3 x 2 − 40 x 3 + L
Constant term = 1 + 3( −4) = −11
3. (1 − x − x 2 )5
= [1 − x (1 + x )]
5
= 1 − 5 x (1 + x ) + 10 x 2 (1 + x )2 −10 x (1 + x ) + L 3
3
= 1 − 5 x − 5 x 2 + 10 x 2 (1 + 2 x + L) −10 x (1 + L) 3
= 1 − 5 x + 5 x 2 + 10 x 3 + L
8. (1 − x + 2 x 2 ) n = [1 − x (1 − 2 x )]n
= 1 − n C1 x (1 − 2 x ) + n C2 x 2 (1 − 2 x )2 + L = 1 − n C1 x (1 − 2 x ) + n C2 x 2 (1 − 4 x + 4 x 2 ) + L = 1 − n C1 x + (2 n C1 + n C2 ) x 2 + L
67
68
Chapter 4 The Binomial Theorem
n(n − 1) 4 x (1 + 2 x 2 n(n − 1)(n − 2) 6 +x2 ) + x (1 + 3 x 6 + L) + L n(n − 1) 4 = 1 + nx 2 + nx 3 + x 2 n(n − 1) 6 + n(n − 1) x 5 + x 2 n(n − 1)(n − 2) 6 + x 6 n(n − 1)(n − 2) 7 + x +L 2 5 Coefficient of x = n(n − 1)
2 n C1 + n C2 = 44 1 2 n + n(n − 1) = 44 2 n 2 + 3n − 88 = 0 (n − 8)(n + 11) = 0 n = 8 or −11 (rejected)
= 1 + nx 2 + nx 3 +
9. (a) (1 − x + 2 x 2 )6
= [1 − x (1 − 2 x )]6 = 1 − 6 x (1 − 2 x ) + 15 x 2 (1 − 2 x )2 − 20 x 3 (1 − 2 x )3 + L = 1 − 6 x + 12 x 2 + 15 x 2 (1 − 4 x + L) − 20 x 3 (1 + L)
Coefficient of x 7 =
= 1 − 6 x + 27 x 2 − 80 3 + L (b) (1 − x + 2 x 2 )6 (1 + x )6
(b)
= (1 − 6 x + 27 x 2 − 80 3 x 3 + L) (1 + 6 x + 15 x 2 + 20 x 3 + L) = 1 + x (27 − 6 × 6 + 15) + x 3 ( −80 + 27 × 6 − 6 × 15 + 20) + L
10. (a) (1 + x − 2 ax 2 ) n
Coefficient of x 5 1 n( n − 1)( n − 2 ) 2
=5
=5 n(n − 1) n−2 =5 2 n = 12
2
= 1 + 6 x 2 + 12 x 3 + L a = 0 , b = 6 , c = 12
Coefficient of x 7
1 n(n − 1)(n − 2) 2
12. (1 + px + qx 2 ) 4 = [1 + x ( p + qx )]4
= [1 + x (1 − 2 ax )]n = 1 + n C1 x (1 − 2 ax )
= 1 + 4 x ( p + qx ) + 6 x 2 ( p + qx )2
+ n C2 x 2 (1 − 2 ax )2 + L = 1 + nx (1 − 2 ax ) 1 + n(n − 1) x 2 (1 + L) + L 2 n(n − 1) 2 = 1 + nx − [2 an − ]x + L 2
= 1 + 4 px + 4 qx 2 + 6 x 2 ( p 2 + 2 pqx + L)
+ 4 x 3 ( p + qx )3 + L + 4 x 3 ( p3 + L) + L = 1 + 4 px + 4 qx 2 + 6 p 2 x 2 + 12 pqx 3 + 4 p3 x 3 + L = 1 + 4 px + 2(3 p 2 + 2 q ) x 2 + 4 p( p 2 + 3q ) x 3 + L
(b) Coefficient of x = n = 7 Comparing the coefficients of x 2 , n(n − 1) −[2 an − ]= 0 2 7×6 =0 2 a(7) − 2 3 a= 2
2(3 p 2 + 2 q ) = 0 ..............(1) 4 p( p 2 + 3q ) = −112 ......(2) By (1), 2 q = −3 p 2 3 q = − p2 2
11. (a) (1 + x 2 + x 3 ) n = [1 + x 2 (1 + x )]n
n(n − 1) 4 x (1 + x )2 2 n(n − 1)(n − 2) 6 + x (1 + x )3 + L 6
−3 2 p )(3)] = −112 2 7 4 p( − p 2 ) = −112 2 −14 p3 = −112
Substitute into (2), 4 p[ p 2 + (
p3 = 8 p=2
= 1 + nx 2 (1 + x ) +
∴
3 q = − ( 4) = −6 2
Chapter 4 The Binomial Theorem
69
4. (1 + 2 x )3 (1 − x )2
13. (a) (1 − tx − x 2 ) 7
= [1 + 3(2 x ) + 3(2 x )2 + (2 x )3 ](1 − 2 x + x 2 )
= [1 − x (t + x )]7 = 1 − 7 x (t + x ) + 21x (t + x ) 2
2
= (1 + 6 x + 12 x 2 + 8 x 3 )(1 − 2 x + x 2 ) = 1 + ( −2 + 6) x + (1 − 12 + 12) x 2 + (8 − 24 + 6) x 3
− 35 x 3 (t + x )3 + L
+ (12 − 16) x 4 + 8 x 5
= 1 − 7tx − 7 x 2 + 21x 2 (t 2 + 2tx + L) − 35 x 3 (t 3 + L) + L
= 1 + 4 x + x 2 − 10 x 3 − 4 x 4 + 8 x 5
= 1 − 7tx − 7 x 2 + 21t 2 x 2 + 42tx 3 − 35t 3 x 3 + L = 1 − 7tx + 7(3t − 1) x 2
2
12 − r ( − x −1 )r 12 Cr (2 x )
−7t (5t 2 − 6) x 3 + L (b) By (a), comparing the coefficients of x 3 , −91t = −7t (5t − 6) 2
2
13t = 5t 2 − 6 5t 2 − 13t − 6 = 0 (5t + 2)(t − 3) = 0 t = 3, −
1 5. (2 x − )12 x The general term is = 12 Cr (2)12 − r ( −1)r x12 − 2 r
It is the constant term when 12 − 2 r = 0 . ∴ r = 6 ∴ The constant term =
12 C6
⋅ 2 6 ( −1)6 = 59 136
x2 2 8 − ) x 2 The general term is
6. ( 2 (rejected) 5
Coefficient of x 2 = 7(27 − 1) = 182 14 − 15. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
x2 8− r 2 r ) ( − ) = 8Cr ( −1)r 2 2 r − 8 x16 − 3r 2 x Term in x 4 when 16 − 3r = 4 , ∴ r = 4 8 Cr (
Coefficient of x 4 = 8 C4 ( −1) 4 2 8 − 8 = 70 7. (1 + x )16 = 1 + 16 C1 x + 16 C2 x 2 + L Coefficient of the 3rd term =
16 C2
= 120
Revision Exercise 4 (p.95) 1 1. (2 x − )5 x 1 1 = (2 x )5 − 5(2 x ) 4 ( ) + 10(2 x )3 ( )2 x x 1 4 1 2 1 3 −10(2 x ) ( ) + 5(2 x )( ) − ( )5 x x x = 32 x 5 − 80 x 3 + 80 x − 40 x −1+ 10 x −3 − x −5
2. ( a − b)3 ( a + b)3
= ( a 2 − b 2 )3 = a 6 − 3a 4 b 2 + 3a 2 b 4 − b 6
8. (2 − x )15 = 215 − 15C1 214 x + L + 15C12 2 3 ( − x )12 + L Coefficient of the 13th term =
15 C12
9. (2 + x + x 3 ) 7
= [2 + x (1 + x 2 )]7 = 2 7 + 7 C1 ⋅ 2 6 x (1 + x 2 ) + 7C2 ⋅ 2 5 x 2 (1 + x 2 )2 + 7C3 ⋅ 2 4 x 3 (1 + x 2 )3 + L = 128 + 448 x (1 + x 2 ) + 672 x 2 (1 + L) + 560 x 3 (1 + L) + L = 128 + 448 x + 672 x 2 + 1 008 x 3 + L 10. (3 + 2 x − x 2 ) 4
1 3. ( x 2 − ) 4 x 1 1 = x 8 − 4( x 2 )3 ( ) + 6( x 2 )2 ( )2 x x 1 4 2 1 3 − 4( x )( ) + ( − ) x x 8 5 2 = x − 4 x + 6 x − 4 x −1 + x −4
= [3 + x (2 − x )]4 = 34 + 4 ⋅ 33 x (2 − x ) + 6 ⋅ 32 x 2 (2 − x )2 + 4 ⋅ 3 x 3 ( 2 − x )3 + L = 81 + 108 x (2 − x ) + 54 x 2 ( 4 − 4 x + L) + 12 x 3 (8 − L) + L = 81 + 216 x + 108 x 2 − 120 x 3 + L
2 3 = 3 640
70
Chapter 4 The Binomial Theorem
It is the constant term when
11. (1 + 2 x ) 4 (1 − x )6 = (1 + 8 x + 24 x 2 + 32 x 3 + L) (1 − 6 x + 15 x 2 − 20 x 3 + L) = 1 + 2 x − 9 x 2 − 12 x 3 + L
12. (1 − 2 x ) 4 (1 + x ) 7 = (1 − 8 x + 24 x 2 + L)(1 + 7 x + 21x 2 + L)
Coefficient of x 2 = 21 − 56 + 24 = −11
1 + 4 x 2 )4 2x 1 = [1 − (1 − 8 x 3 )]4 2x 1 1 = 1 − 4( )(1 − 8 x 3 ) + 6( )2 (1 − 8 x 3 )2 2x 2x 1 3 1 3 3 − 4( ) (1 − 8 x ) + ( ) 4 (1 − 8 x 3 ) 4 2x 2x Constant term = 1 + 12 = 13
13. (1 −
14. (1 + x − 2 x 2 )9 (1 + x ) 4
= [1 + x (1 − 2 x )]9 (1 + x ) 4 = [1 + 9 x (1 − 2 x ) + 36 x 2 (1 − 2 x )2 + 84 x 3 (1 − 2 x )3 + L](1 + 4 x + 6 x 2 + 4 x 3 + L) = [1 + 9 x − 18 x 2 + 36 x 2 (1 − 4 x + L) + 84 x 3 (1 + L) + L](1 + 4 x + 6 x 2 + 4 x 3 + L) = (1 + 9 x + 18 x 2 − 60 x 3 )(1 + 4 x + 6 x 2 + 4 x 3 + L) Coefficient of x 3 = 4 + 9 × 6 + 18 × 4 − 60 = 70 15. (1 + 2 x ) n = 1 + 2 nx +
1 n(n − 1)(2 x )2 2
1 + n(n − 1)(n − 2)(2 x )3 6 1 + n(n − 1)(n − 2)(n − 3)(2 x ) 4 + L 24 Q Coefficient of x 3 = Coefficient of x 4 ∴
2 3 ⋅ n(n − 1)(n − 2) 2 4 ⋅ n(n − 1)(n − 2)(n − 3) = 6 24 4 2 = (n − 3) 3 3 2 = n−3 n=5
16. The general term in the expansion 1 1 (2 x 2 + ) n = n Cr (2 x 2 ) n − r ( )r 2x 2x = n Cr ⋅ 2 x n − 2 r x 2 n − 3r ∴ The 7th term = n C6 ⋅ 2 n −12 x 2 n −18
2 n − 18 = 0 . ∴ n = 9
∴ The value of the 7th term 21 = 9 C6 ⋅ 2 −3 = 2 17. The general term in the expansion a a ( x 2 + ) 7 = 7Cr ( x 2 ) 7 − r ( )r 2x 2x a = 7Cr ( )r x14 − 3r 2 a 21 2 ∴ A8 = 7C2 ( )2 = a 2 4 a 7 A11 = 7C1 ( ) = a 2 2 A8 = 6 A11 21 2 7 a = 6( )a 4 2 a( a − 4 ) = 0 a = 4 or 0 (rejected) 2 n 2 ) = ( ax ) n + n( ax ) n −1 ( 2 ) x2 x n(n − 1) 2 + ( ax ) n − 2 ( 2 )2 + L 2 x n(n − 1) 4 The third term = ( ax ) n − 2 ( 4 ) 2 x = 2 n(n − 1)a n − 2 x n − 6
18. ( ax +
Q The third term is independent of x. ∴
n − 6 = 0 , i.e. n = 6
The coefficient of the third term: 15 4 4 15 2(6)(5)a = 4 1 4 a = 16 1 1 (rejected) a = − or 2 2
2 n(n − 1)a n − 2 =
19. (1 + 2 x )5 (1 − x ) n = [1 + 5(2 x ) + 10(2 x )2 + L] n(n − 1) 2 [1 − nx + x + L] 2 = (1 + 10 x + 40 x 2 + L) n2 − n 2 x + L) 2 (a) Coefficient of x = − n + 10 = 1 n =9 (1 − nx +
Chapter 4 The Binomial Theorem
n2 − n − 10 n + 40 2 92 − 9 = − 90 + 40 2 = −14
(b) Coefficient of x 2 =
20 − 22. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons. x x n(n − 1) x 2 23. (1 + ) n = 1 + n( ) + ( ) 2n 2n 2 2n x 3 1 + n(n − 1)(n − 2)( ) L 6 2n Coefficient of x 2 = n(n − 1) ( 1 )2 2 2n n −1 8n 2n n
1 10 1 = 10 = 10 =5 =
1 1 n(n − 1)(n − 2)( )3 6 2n 1 1 3 = ×5× 4×3×( ) 6 10 1 = 100
Coefficient of x 3 =
24. (a) (1 + x + px 2 ) q = [1 + x (1 + px )]q = 1 + qx (1 + px ) +
1 q(q − 1) x 2 (1 + px )2 2
1 q(q − 1)(q − 2) x 3 (1 + px )3 + L 6 1 = 1 + qx + pqx 2 + q(q − 1) x 2 (1 + 2 px 2 1 + L) + q(q − 1)(q − 2) x 3 (1 + L) + L 6 q(q − 1) 2 x = 1 + qx + pqx 2 + 2 q(q − 1)(q − 2) 3 + pq(q − 1) x 3 + x +L 6 q(q − 1) 2 = 1 + qx + [ pq + ]x 2 q(q − 1)(q − 2) 3 +[ pq(q − 1) + ]x + L 6 1 = 1 + qx + q(2 p + q − 1) x 2 2 1 + q(q − 1)(6 p + q − 2) x 3 + L 6 +
71
q = 6 ...............................(1) (b) 1 2 q(2 p + q − 1) = 27 .......(2) Substitute q = 6 into (2). 1 (6)(2 p + 6 − 1) = 27 2 2p + 5 = 9 p=2 1 (6)(5)(12 + 6 − 2) 6 = (5)(16) = 80
Coefficient of x 3 =
25. (a) (1 − px )6 − (1 + x ) n = (1 − 6 px + 15 p 2 x 2 + L) 1 − [1 + nx + n(n − 1) x 2 + L] 2 1 = ( −6 p − n) x + [15 p 2 − n(n − 1)]x 2 + L 2 1 2 = −(6 p + n) x + (30 p − n 2 + n) x 2 + L 2 (b) −(6 p + n) = −17 .................(1) 1 (30 p 2 − n 2 + n) = 50 .......(2) 2 17 − n By (1), p = ..............(3) 6 Put (3) into (2), 1 17 − n 2 [30( ) − n 2 + n] = 50 2 6 5(289 − 34n + n 2 ) − 6n 2 + 6n = 600 n 2 + 164n − 845 = 0 n=5
or
−169 (rejected)
Put n = 5 into (3), p=
17 − 5 =2 6
Enrichment 4 (p.97) b 1. (a) ( ax + ) n x b b = ( ax ) n + n C1 ( ax ) n −1 ( ) + n C2 ( ax ) n − 2 ( )2 x x n −3 b 3 n−4 b 4 + n C3 ( ax ) ( ) + n C4 ( ax ) ( ) +L x x = a n x n + n C1a n −1bx n − 2 + n C2 a n − 2 b 2 x n − 4 + n C3 a n − 3b 3 x n − 6 + n C4 a n − 4 b 4 x n − 8 + L
72
Chapter 4 The Binomial Theorem
(b) The fifth term is the constant term. n−8 = 0 n=8 2. (a) (1 + 3 x ) m + (1 + 5 x ) n m( m − 1) = [1 + m(3 x ) + (3 x )2 + L] + 2 n(n − 1) [1 + n(5 x ) + (5 x )2 + L] 2 = 2 + (3m + 5n) x 9m( m − 1) 25n(n − 1) 2 +[ + ]x + L 2 2 Comparing coefficients of x, 3m + 5n = 19 Q m and n are positive integers.
∴
n(n − 1)(n − 2) 7n(n − 1) + + 7n − 49 6 2 (3)(2) 7(3)(2) = + + 7(3) − 49 6 2 = −6
b=
4. Let the three consecutive coefficients be the coefficients of the (r − 1) th, rth and (r + 1) th terms, then n Cr − 2
= 3a ................................(1)
n Cr −1
= 12 a ..............................(2)
n Cr
= 28a .............................. (3)
(1) , (2)
m = 3, n = 2
9m( m − 1) 25n(n − 1) + 2 2 9(3)(2) 25(2) = + 2 2 = 52
(b) a =
3. (a) (1 + x − 2 x 2 ) 7 = [1 + x (1 − 2 x )]
7
= 1 + 7 x (1 − 2 x ) + 21x 2 (1 − 2 x )2 + 35 x 3 (1 − 2 x )3 + L = 1 + 7 x − 14 x + 21x (1 − 4 x + L) 2
2
+ 35 x 3 (1 + L) + L = 1 + 7 x + 7 x − 49 x + L 2
(b) (1 + x − 2 x ) (1 + x ) 2 7
3
7 + n = 10 n=3
n(n − 1) + 7n + 7 2 (3)(2) = + 7(3) + 7 2 = 31
3a 12 a n Cr −1 3 r −1 = n − r + 2 12 15r − 3n = 18 ..................(4) =
n Cr −1
12 a = 28a C n r 3 r = n − r +1 7 10 r − 3n = 3 .....................(5)
(4) − (5) : 5r = 15 r=3 Put r = 3 into (4),
15(3) − 3n = 18 n=9 Put n = 9 and r = 3 into (3), = 28a a=3
9 C3
n
= (1 + 7 x + 7 x 2 − 49 x 3 + L)[1 + nx + n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + L] 2 6 n(n − 1) = 1 + ( 7 + n) x + [ + 7n + 7]x 2 2 n(n − 1)(n − 2) 7n(n − 1) +[ + + 7n − 49]x 3 6 2 +L By comparing coefficients,
a=
(2) , (3)
n Cr − 2
Classwork 1 (p.84) 1. (a) 5!= 5 × 4 × 3 × 2 × 1 = 120 (b)
(c)
9! 9 × 8 × 7! = 7! 7! = 72 7 C3
7! 3!(7 − 3)! 7! = 3! 4! 7 × 6 × 5 × 4! = 3! 4! = 35 =
Chapter 4 The Binomial Theorem
2. (a)
(b)
n Cn − 3
= = = =
(c)
∴ The coefficient of x 3 = 7C3 ( −2)3 = −280
n! 3!(n − 3)! n(n − 1)(n − 2)n − 3! = 6(n − 3)! 1 = n(n − 1)(n − 2) 6 =
n C3
n + 2 C2
(b) The general term in the expansion = 6 Cr ( x 2 )6 − r (2)r ∴ The term in x 6 = 6 C3 ( x 2 )3 (2)3
n! (n − 3)![n − (n − 3)]! n! (n − 3)! 3! n(n − 1)(n − 2)(n − 3)! (n − 3)! 3! 1 n(n − 1)(n − 2) 6
(n + 2)! 2!(n + 2 − 2)! (n + 2)! = 2! n! (n + 2)(n + 1)n! = 2! n! 1 = (n + 1)(n + 2) 2
∴ The coefficient of x 6 = 160 (c) The general term in the expansion 1 = 6 Cr (3 x )6 − r ( )r x = 6 Cr ⋅ 36 − r x 6 − 2 r It is the constant term when
6 − 2r = 0 . ∴ r = 3 ∴ The constant term = 6 C3 ⋅ 33 = 540 (d) The general term in the expansion
=
2 = 9 Cr ( x 2 )9 − r ( − )r x 2 18 − 2 r = 9 Cr ( x )( − )r x = 9 Cr ( −2)r x18 − 3r
It is the constant term when 18 − 3r = 0 . ∴ r = 6 ∴ The constant term = 9 C6 ( −2)6 = 5 376
Classwork 2 (p.89) 1. (a) (1 + 2 a) 4 = 1+ +
4 C1 (2 a) + 4 C2 (2 a) 3 4 4 C3 (2 a) + (2 a) 2 3
Classwork 3 (p.90)
2
= 1 + 8a + 24 a + 32 a + 16 a
1. (a) ( x 2 − 1)9
4
= ( −1 + x 2 )9 = ( −1)9 + 9( −1)8 ( x 2 ) + 36( −1) 7 ( x 2 )2 + L
(b) (3 − b) 4 = (3) 4 + +
3 2 2 4 C1 (3) ( − b ) + 4 C2 (3) ( − b ) 3 4
4 C3 (3)( − b )
+ ( − b)
= 81 − 108b + 54b − 12 b + b 2
(c) ( x + 2 y)
73
3
4
3
= ( x )3 + 3C1 ( x )2 (2 y) + 3C2 ( x )(2 y)2 + (2 y)3 = x 3 + 6 x 2 y + 12 xy 2 + 8 y 3
(d) ( y − 3 x )3 = ( y)3 + 3C1 ( y)2 ( −3 x ) + 3C2 ( y)( −3 x )2 + ( −3 x )3
= y 3 − 9 y 2 x + 27 yx 2 − 27 x 3 2. (a) The general term in the expansion = 7Cr ( −2 x )r ∴ The term in x 3 = 7C3 ( −2 x )3
= −1 + 9 x 2 − 36 x 4 + L 1 (b) (2 + ) 4 x 1 1 = (2) 4 + 4(2)3 ( ) + 6(2)2 ( )2 x x 1 3 1 4 + 4(2)( ) + ( ) x x 32 24 8 1 = 16 + + + + x x2 x3 x 4 1 (c) ( x 2 − 1)9 (2 + ) 4 x = ( −1 + 9 x 2 − 36 x 4 + L) 32 24 8 1 + 2 + 3 + 4) (16 + x x x x ∴ The constant term in the expansion = ( −1)(16) + 9(24) + ( −36)(1) = 164
74
Chapter 4 The Binomial Theorem
2. (3 x 2 + 1) n = (1 + 3 x 2 ) n = 1 + n C1 (3 x 2 ) + n C2 (3 x 2 )2 + n C3 (3 x 2 )3 + L
Compare coefficient to a + bx 2 + cx 4 + dx 6 + L b = 3 ⋅ n C1 , d = 27 ⋅ n C3 Q
d = 108b
27 ⋅ n C3 = 108(3 ⋅ n C1 ) n! n! 27 ⋅ = 324 ⋅ 3!(n − 3)! (n − 1)! 9 (n − 1)(n − 2) = 324 2 n 2 − 3n + 2 = 72 n 2 − 3n − 70 = 0 (n + 7)(n − 10) = 0 n = 10
∴ Comparing the coefficients of x, n C1 p = −7 np = −7 −7 p= ......................(1) n Comparing the coefficients of x 2 , 2 ⋅ n C1 + p 2 ⋅ n C2 = 35 1 2 n + p 2 ⋅ n(n − 1) = 35 ..........(2) 2 Put (1) into (2), 2n + (
or −7 (rejected)
Classwork 4 (p.92) 1. (1 + x − 3 x 2 )6 = [1 + x (1 − 3 x )]6 = 1 + 6 x (1 − 3 x ) + 15 x 2 (1 − 3 x )2 + 20 x 3 (1 − 3 x )3 + L = 1 + 6 x − 18 x 2 + 15 x 2 (1 − 6 x + L) + 20 x 3 (1 + L) + L = 1 + 6 x − 18 x 2 + 15 x 2 − 90 x 3 + 20 x 3 + L = 1 + 6 x − 3 x 2 − 70 x 3 + L 1 2. (1 + x − 3 x 2 )6 (1 − )3 x = (1 + 6 x − 3 x 2 − 70 x 3 + L) 1 1 1 [1 + 3( − ) + 3( − )2 + ( − )3 ] x x x = (1 + 6 x − 3 x 2 − 70 x 3 + L) 3 3 1 (1 − + 2 − 3 ) x x x
∴ The constant term = 1 + 6( −3) − 3(3) − 70( −1) = 44
Classwork 5 (p.93) (1 + px + 2 x 2 ) n = [1 + x ( p + 2 x )]n = 1 + n C1 x ( p + 2 x ) + n C2 x 2 ( p + 2 x )2 + L = 1 + n C1 px + 2 n C1 x 2 + n C2 x 2 ( p 2 + L) + L = 1 + n C1 px + (2 ⋅ n C1 + p 2 ⋅ n C2 ) x 2 + L
−7 2 1 ) n(n − 1) = 35 n 2 49 n − 1 2n + ⋅ = 35 2 n 4n 2 + 49n − 49 = 70 n 4n 2 − 21n − 49 = 0 (n − 7)( 4n + 7) = 0 n=7
Put n = 7 into (1), p = −
or 7 = −1 7
−
7 (rejected) 4