若{ an}为等差数列,设公差为d,则
1*a1+2*a2+3*a3+……+n*an=(1+2+……+n)*a1+[1*2+2*3+……+(n-1)n]d
=n(n+1)a1/2+(1^2+2^2+3^2+……+n^2)d-(1+2+……+n)d
=n(n+1)a1/2+[n(n+1)(2n+1)d/6-n(n+1)d/2]
=n(n+1)a1/2+(n-1)n(n+1)d/3
a1+a2+a3+……+an=s=na1+n(n-1)d/2
1*a1+2*a2+3*a3+……+n*an=2(n+1)s/3-n(n+1)a1/6
等比数列就不求了,你应该结合具体情况的
f(x)=a0+a1x+a2x^2+..+anx^n, f(1)=a0+a1+..+an=S-a0
f'(x)=a1+2a2x+3a3x^2+..nanx^(n-1)
f'(1)=a1+2a2+..+nan
适用于很多时候