Math and Stat,binomal易到喊

(b) 1/(1-x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...
1/(1-x)^4 = [(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...)^2]^2
(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...)^2 =
x + x^2 + x^3 + x^4 + x^5 + x^6 + ... +
x^2 + x^3 + x^4 + x^5 + x^6 + ... +
x^3 + x^4 + x^5 + x^6 + ... +
x^4 + x^5 + x^6 + ... +
x^5 + x^6 + ... +
x^6 + ...
=(1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^4 + 7x^6 + ...)
The coefficient of x^k in the above expression is (k+1)

1/(1-x)^4 = [(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ...)^2]^2
=(1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^4 + 7x^6 + ...)^2

The coefficient of x^r is formed by
1 * (r+1)x^r +
2x * (r)x^(r-1) +
3x^2 * (r-1)x^(r-2) +
4x^3 * (r-2)x^(r-3) + ... +
(r+1)x^r * 1

The coefficient of x^8 in the expansion is
1 * (r+1) + 2 * (r) + 3 * (r-1) + 4 * (r-2) + ... + (r+1) * 1
= 1 * [(r+2)-1] + 2 * [(r+2)-2]) + 3 * [(r+2)-3] + 4 * [(r+2)-4] + ... + (r+1) * [(r+2)-(r+1)]
=[1+2+3+...+(r+1)]*(r+2) - [1*1+2*2+3*3+...+(r+1)*(r+1)]
=(r+1)(r+2)(r+2)/2-(r+1)(r+2)[2(r+1)+1]/6
=(r+1)(r+2)[3(r+2)-(2r+3)]/6
=(r+1)(r+2)(3r+6-2r-3)/6
=(r+1)(r+2)(r+3)/6

2009-06-15 22:51:15 補充：
Correction:
The coefficient of x^r in the expansion is
.........
=(r+1)(r+2)(r+3)/6

2009-06-16 08:46:19 補充：
I don't know what level of question this one is. But 藍閃蝶 is correct that differentiation is simpler.
(1-x)^-1=Sum(n=0 to infinity)x^n
Differentiate both side wrt x
(1-x)^-2=Sum(n=1 to infinity)nx^(n-1)

2009-06-16 08:46:26 補充：
again
2(1-x)^-3=Sum(n=2 to infinity)n(n-1)x^(n-2)
again
6(1-x)^-4=Sum(n=3 to infinity)n(n-1)(n-2)x^(n-3)
(1-x)^-4=1/6Sum(n=3 to infinity)n(n-1)(n-2)x^(n-3)
Coefficient of x^r term is when n-3=r ot n=r+3
Hence the coefficient is (r+3)(r+3-1)(r+3-2)/6 = (r+1)(r+2)(r+3)/6

b) Using differentiation is simpler