M1 ~~~~ help THX a lot!!!

[匿名]

2013-12-30 12:10 am

1) Find the coefficient of x^2 y^6 in the expansion of (5x-y)^8

2)It is given that (2+x/10)^n =1024 + ax +bx^2 + term involving higher power of x

(a) Find the value of n
(b) Find the value s of a and b

3) (a) Expand (1-bx)^4 in ascending powers of x

(b) Consider the expansion (2+x) ^3 (1-bx) ^4
(i) Express the coefficient of x^2 in the expansion in terms of b
(ii) If the coefficient of x^2 in the expansion is -3 , find the value of b.

Thanks a lot !!!!!!!!!!!!!!!

回答 (1)

fooks

2013-12-30 1:14 am

✔ 最佳答案

(1)
Let (n + 1)th term be the x²y⁶ term.

x⁸⁻ⁿyⁿ = x²y⁶
Hence, n = 6

The coefficient of x²y⁶
= 8C6 * (5)² * (-1)⁶
= (8!/6!2!) * 25
= 28 * 25
= 700

(2)
(a)
Constant term :
2ⁿ = 1024
2ⁿ = 2¹⁰
n = 10

(b)
a = 10C1 * 2⁹ * (1/10)
a = 10 * 512 * (1/10)
a = 512

b = 10C2 * 2⁸ * (1/10)²
b = (10!/2!8!) * 256 * (1/100)
b = 45 * 256 * (1/100)
b = 576/5

(3)
(a)
(1 - bx)⁴
= 1 - 4C1(bx) + 4C2(bx)² - 4C3(bx)³ + (bx)⁴
= 1 - 4bx + 6b²x² - 4b³x³ + b⁴x⁴

(b)(i)
(2 + x)³
= 2³ + 3C1*2²*x + 3C2*2*x² + x³
= 8 + 12x + 6x² + x³

(2 + x)³ (1 - bx)⁴ = (8 + 12x + 6x² + ...)(1 - 4bx + 6b²x² + ......)

The x² term in expansion of (2 + x)³ (1 - bx)⁴
= 8*6b² + 12*(-4b) + 6*1
= 48b² + 48b + 6

(b)(ii)
48b² + 48b + 6 = -3
48b² + 48b + 9 = 0
16b² + 16b + 3 = 0
(4b + 1)(4b + 3) = 0
b = -1/4 or b = -3/4

參考: fooks

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