F4 Binomial Expansion

(a) (x + 1)^n
= nC0 + nC1x + nC2x^2 + ...
= 1 + nx + [n(n-1)/2]x^2 + ...
(b) (x+1)^n(1-x+2x^2)
= (1 + nx + [n(n-1)/2]x^2 + ...)(1-x+2x^2)
= 1 + nx + [n(n-1)/2]x^2 + ... - x - nx^2 + ... + 2x^2 + ...
= 1 +(n-1)x + [n(n-1)/2 - n + 2]x^2 + ...
Comparing coefficients, n-1 = 8 => n = 9
n(n-1)/2 - n + 2 = k
k = (9)(8)/2 - 9 + 2 = 29

2009-12-17 20:28:08 補充：
(1 + nx + [n(n-1)/2]x^2 + ...)(1-x+2x^2)
= (1 + nx + [n(n-1)/2]x^2 + ...) -x(1 + nx + [n(n-1)/2]x^2 + ...) +2x^2(1 + nx + [n(n-1)/2]x^2 + ...)
= 1 + nx + [n(n-1)/2]x^2 + ... -x - nx^2 + ... + 2x^2 + ...

2009-12-17 20:44:04 補充：
x^3或以上都不用理會嘛,所以不用乘,不用表達出來,用"..."代表就OK

2009-12-17 20:55:30 補充：
這是中四可以明白的,沒有太大難度.
(1 + nx + [n(n-1)/2]x^2 + ...) 乘(1- x + 2x^2)
將第二個括號的1, -x 及2x^2 續個去乘1 + nx + [n(n-1)/2]x^2 + ...
1乘1 + nx + [n(n-1)/2]x^2 + ...得1 + nx + [n(n-1)/2]x^2 + ...
-x乘1 + nx + [n(n-1)/2]x^2 + ...得-x - nx^2 + 第三項是x三次方,這道題不用理會
2x^2乘1 + nx + [n(n-1)/2]x^2 + ...得2x^3 + 第二項是x三次方,第三項是x4次方,不用理會

2009-12-17 21:42:00 補充：
The question says coefficient of the x term is 8
The expansion shows that the coefficient of the x term is n - 1
So 8 = n - 1 => n = 9
The question says coefficient of the x^2 term is k
The expansion indicates that the coefficient of the x^2 term is n(n-1)/2 - n + 2, and
so k = (9)(9-1)/2 - 9 + 2 = 29