M2 二項式定理

1.
(1-2x)^6 + (2+kx)^4
= [1 + 6C1[-2x] + 6C2[-2x]^2 + 6C3[-2x]^3 + ... ] + [16 + 4C1[2^3][kx] + 4C2[2^2][kx]^2 + 4C3[2][(kx)^3] + ... ]
Coefficient of x^3 = 6C3[-8] + 4C3[2][(k^3]
= -160 + 8k^3 = -159
8k^3 = 1
k = 1/2

2.
[(x-1)^7] [(2+3x)^5]
= [(-1)^7 + 7C1[x][(-1)^6] + 7C2[x^2][(-1)^5] +...] [32 + 5C1[2^4][3x] + 5C2[2^3][(3x)^2] +... ]
= [-1 + 7x - 21x^2 +...] [32 + 240x + 720x^2 +... ]
= [-1 + 7x - 21x^2 +...] [32 + 240x + 720x^2 +... ]
the coefficient of x^2:
(-1)(720) + (7)(240) + (-21)(32) = 288

3.
(1+qx)^n = 1 - 22x + 55q^2x^2 + ...
-22 = nC1[q] => nq = -22
55q^2 = nC2[q^2] => n(n-1)/2 = 55
n^2 - n - 110 = 0 => (n-11)(n+10) = 0 => n = 11 or -10 (neglected)
q = -2 and n = 11

4.
(1+2x-3x^2)^3 = [(1+3x)(1-x)]^3 = [(1+3x)^3] [(1-x)]^3]
= [1 + 3(3x) + 3(3x)^2 + (3x)^3] [1 + 3(-x) + 3(-x)^2 + (-x)^3]
= [1 + 9x + 9x^2 + 27x^3] [1 - 3x + 3x^2 - x^3]
= 1 + 6x - 15x^2 + 26x^3 + ...

5a.
(1+x-2x^2)^4 = [(1+2x)(1-x)]^4 = [(1+2x)^4] [(1-x)]^4]
= [1 + 4(2x) + 6(2x)^2 + 4(2x)^3 + ...] [1 + 4(-x) + 6(-x)^2 + 4(-x)^3 + ...]
= [1 + 8x + 24x^2 + 32x^3 + ...] [1 - 4x + 6x^2 - 4x^3 + ...]
= 1 + 4x - 2x^2 - 20x^3 + ...
5b.
(1-3x)^3 = [1 + 3(-3x) + 3(-3x)^2 + (-3x)^3]
= [1 - 9x + 27x^2 - 27x^3]
[(1+x-2x^2)^4] [(1-3x)^3]
= [1 + 4x - 2x^2 - 20x^3 + ...] [1 - 9x + 27x^2 - 27x^3]
= 1 - 5x - 11x^2 + 79x^3
thus, a = -5, b = -11, c = 79

圖片參考：http://i42.tinypic.com/2mdotpf.png


http://i42.tinypic.com/2mdotpf.png

2011-12-02 20:57:28 補充：
錯左好多,更正:http://i41.tinypic.com/25us9xy.png

2011-12-02 20:58:10 補充：
karf第4題好似錯左