数列通项公式为n*2^n,求数列前n项和

数列通项公式为n*2^n,求数列前n项和
数列通项公式为n*2^n,求数列前n项和

数列通项公式为n*2^n,求数列前n项和
S(n)=(n-1)×2^(n+1)+2
解法一：
S(n)=2^1+2×2^2+3×2^3+…+n×2^n
=n×(2^1+2^2+2^3+…+2^n)-[2^1+2^2+2^3+…+2^(n-1)]-[2^1+2^2+2^3+…+2^(n-2)]-…-(2^1+2^2)-(2^1)-0
=n×[2^(n+1)-2]-{(2^n-2)+[2^(n-1)-2]+…+(2^3-2)+(2^2-2)+(2^1-2)}
=n×2^(n+1)-2×n-{[2^n+2^(n-1)+…+2^3+2^2+2^1]-2×n}
=n×2^(n+1)-2×n-[2^(n+1)-2-2×n]
=n×2^(n+1)-2^(n+1)+2
=(n-1)×2^(n+1)+2
解法二：
S(n)=2^1+2×2^2+3×2^3+…+(n-1)×2^(n-1)+n×2^n
2×S(n)=2^2+2×2^3+3×2^4+…+(n-1)×2^n+n×2^(n+1)
S(n)=2×S(n)-S(n)
=2^2+2×2^3+3×2^4+…+(n-1)×2^n+n×2^(n+1)-[2^1+2×2^2+3×2^3+…+(n-1)×2^(n-1)+n×2^n]
=n×2^(n+1)-(2^1+2^2+2^3+…+2^n)
=n×2^(n+1)-[2^(n+1)-2]
=(n-1)×2^(n+1)+2

错位相减
Sn=1·2^1+2·2^2+3·2^3+····+n·2^n
2Sn= 1·2^2+2·2^3+····+（n-1)·2^n+n·2^(n+1)
两式相减
-Sn=2-n·2^(n+1)+(2^2+2^3+···+2^n)
Sn=n·2^(n+1)-2+4[1-2^（n-1)]

Sn=2+2·2²+3·2³+...+n·2^n
2Sn=2²+2·2³+3·2^4+…+(n-1)·2^n+n·2^(n+1)
2Sn-Sn=Sn=-2-2²-2³-2^4-...-2^n+n·2^(n+1)
=2+(n-1)·2^(n+1)