级数∑(n=1,∞)x^(n+1)=x^2/(1-x)=-1-x+1/(1-x)两边求导： ∑(n=1,∞)(n+1)x^(n)=x^2/(1-x)=-1+1/(1-x)^2再求导： ∑(n=1,∞)n(n+1)x^(n-1)=x^2/(1-x)=2/(1-x)^3所以：∑(n=1,∞)n(n+1)x^(n)=2x/(1-x)^3 |x|<1
