已知an=n×2缠,求前n项和Sn

Sn=a1+a2++an=(2+1-3×1)+(2²+2-3×2²)++(2ⁿ+n-3×n²)=(2+2²++2ⁿ)+(1+2++n)-3×(1²+2²++n²)=2×(2ⁿ-1)/(2-1)+n(n+1)/2-3×[n(n+1)(2n+1)/6]=2^(n+1)-2+n(n+1)/2-n(n+1)(2n+1)/2=2^(n+1)-2+[n(n+1)/2](1-2n-1)=2^(n+1)-2-2n²(n+1)/2=2^(n+1)-n³-n²-2^表示指数，2^(n+1)表示2的n+1次方。提示：用到三个公式：等比数列求和公式：Sn=a1(qⁿ-1)/(q-1)，本题中a1=2q=2；1+2++n=n(n+1)/21²+2²++n²=n(n+1)(2n+1)/6

Sn=1x2+2x2^2+3x2^3+……+nx2^n
2Sn= 1x2^2+2x2^3+……+(n-1)x2^n+nx2^(n+1)
上式-下式得：
-Sn=1x2+2^2+2^3+……+2^n-nx2^(n+1)
=2(1-2^n)/(1-2)-nx2^(n+1)
=-2+2^(n+1)-nx2^(n+1)
=-2-(n-2)x2^(n+1)
则Sn=2+(n-2)x2^(n+1)

let
S=1.2^1+2.2^2+...+n.2^n (1)
2S= 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) -(2^1+2^2+...+2^n)
= n.2^(n+1) -2(2^n -1)
= 2 + (2n-2).2^n
an = n.2^n
Sn = a1+a2+...+an =S =2 + (2n-2).2^n