Question

1/1+x展开为x-3的幂级数

Answer 1

f(x) = 1/(1+x) =1/(4+x-3) = (1/4){1/[1+(x-3)/4]}
= (1/4)∑<n=0,∞>(-1)^n*[(x-3)/4]^n
= ∑<n=0,∞>(-1)^n*(x-3)^n/4^(n+1).
收敛区间 -1<(x-3)/4<1, -4<x-3<4, 得 -1<x<7.