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