1/(x*2-1)展开成关于x幂级数
问题描述：
1/(x*2-1)展开成关于x幂级数
答
|x|无穷}x^n,
1/(1+x) = Sum_{n=0->无穷}(-x)^n,
1/(x^2-1) = (1/2)[1/(x-1) - 1/(x+1)] = -1/2[ 1/(1-x) + 1/(1+x) ]
= Sum_{n=0->无穷}(-1/2)[1 + (-1)^n]x^n
= Sum_{n=0->无穷} {[(-1)^(n-1) - 1]/2}*x^n
|x|