高等数学，微积分，幂级数求和函数

2025-05-19 03:33:55

推荐回答（2个）

回答1：

f(x)=1*2x+2*3x²+...
=(1*x²+2*x³+...)'
=[x(x+2x²+3x³+...)]'
而我们知道x+2x²+3x³+...=x/(1-x)²
所以f(x)=[x²/(1-x)²]'
求导就自己写了

回答2：

S(x) = 1·2x + 2·3x^2 + 3·4x^3+ ......
= ∑n(n+1)x^n = [∑nx^(n+1)]'
= [∑(n+2)x^(n+1) - ∑2x^(n+1)]'
= {[∑x^(n+2)]' - 2x^2/(1-x)}' ( |x| < 1 )
= {[x^3/(1-x)]' - 2x^2/(1-x)}' = {[-x^2-x-1+1/(1-x)]' + 2x+2-1/(1-x)}'
= {-2x -1 + 1/(1-x)^2 + 2x + 2 - 1/(1-x)}'
= [1 + 1/(1-x)^2 - 1/(1-x)]' = 2/(1-x)^3 - 1/(1-x)^2
= (1+x)/(1-x)^3 ( |x| < 1 )