求数列1×2.3×4.5×8.（2n－1）×2^n的前n项和

数列为{an},前n项和Sn
an=(2n-1)×2ⁿ
Sn=a1+a2+a3+...+an=1×2+3×2²+5×2³+...+(2n-1)×2ⁿ
2Sn=1×2²+3×2³+...+(2n-3)×2ⁿ+(2n-1)×2^(n+1)
Sn-2Sn=-Sn=2+2×2²+2×2³+...+2×2ⁿ -(2n-1)×2^(n+1)
=2(2+2²+...+2ⁿ) -(2n-1)×2^(n+1) -2
=2×2×(2ⁿ-1)/(2-1) -(2n-1)×2^(n+1) -2
=(3-2n)×2^(n+1) -6
Sn=(2n-3)×2^(n+1) +6