Question

一道幂级数展开的题目求解

展开成等式的不太明白怎么写，麻烦给出具体过程！

Answer 1

f(x) = 1/(x^2-4x+3) = (1/2)[1/(1-x) - 1/(3-x)]
= (1/2){1/[2-(x+1)] - 1/[4-(x+1)]}
= (1/4)/[1-(x+1)/2] - (1/8)/[1-(x+1)/4]
= (1/4)∑<n=0,∞>[(x+1)/2]^n - (1/8)∑<n=0,∞>[(x+1)/4]^n
= ∑<n=0,∞>[1/2^(n+2)-1/2^(2n+3)](x+1)^n
收敛域 -1<(x+1)/2<1, -1<(x+1)/4<1, 解得 -3<x<1。