求x+2x^2+3X^3+……+nx^n的前n项和Sn

当x≠1时
Sn=x+2x^2+3X^3+……+nx^n
xSn=x^2+2x^3+nx^(n+1)
相减得
(1-x)Sn=x+x^2+x^2+...+x^n-nx^(n+1)
=[x^(n+1)-x]/(x-1)-nx^(n+1)
Sn=-[x^(n+1)-x]/(x-1)^2-nx^(n+1)/(1-x)
当x=1时
Sn=n(n+1)/2

Sn=x+2x^2+3X^3+……+nx^n ①
xSn= x^2+2X^3+……+(n-1)x^n+nx^(n+1) ②
②-①，得 (x-1)Sn=nx^(n+1)-(x+x^2+x^3+…+x^n)
∴ x=0，Sn=0；x=1，Sn=n(n+1)/2；
x≠1,0时，Sn=nx^(n+1)/(x-1)-x(x^n-1)/(x-1)²