若是，则记 f(x)=∑ x^(n+2)=x^2+x^3+x^4+...=x^2/(1-x) (-1得 f'(x)=∑(n+2)x^(n+1), f''(x)=∑(n+1)(n+2)x^n, 于是，幂级数 ∑(n+1)(n+2)x^n 的和函数是 g(x)=f''(x)=[1/(1-x)-(1+x)]''=[1/(1-x)^2-1]'=2/(1-x)^3. (-1
