【急.已知数列{(2n-1)·2^n},求其前N项和Sn利用错项相减求出Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1所以Sn =6-2^(n+2)*(2-n)为什么再代入值验算时都不对,

有正解了，飘过

-Sn=2+【2^3+2^4+...+2^(n+1)】- (2n-1)*2^(n+1) 中括号内有n-1项的等比数列求和
= 2+ 8【2^(n-1)-1】- (2n-1)*2^(n+1)
= -6-(2n-3)2^(n+1)
Sn = 6 + (2n-3)2^(n+1)

这个式子是对的Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1然后是这样计算的：-Sn=2+【2^3+2^4+...+2^(n+1)】- (2n-1)*2^(n+1) 中括号内有n-1项的等比数列求和= 2+ 8【2^(n-1)-1】- (2n-1)*2^(n+1) = -6-(2n-3)2^(n...

an = n2^(n+1) - 2^n
令bn = n2^(n+1)，cn = 2^n
bn的前n项为Bn，cn的前n项为Cn，an的前n项为Sn,
Bn = 1*2^2+2*2^3+3*2^4+......+n2^(n+1)
2Bn= 1*2^3+2*2^4+.......+(n-1)2^(n+1)+n2^(n+2)
相减：
-Bn = 2^2+2^3+.......+2^(n+1)-n2^(n+2)
=4(2^n - 1)-n2^(n+2)
Bn = (n-1)2^(n+2) + 4
Cn = 2^(n+1)-2
Sn = Bn + Cn
=(n-1)2^(n+2) +2^(n+1) + 2