已知an 求Sn（数列求和问题,要求用错位相减法）

1.Sn=1×2+2×2^2+3×2^3+.+n×2^n (1)
2Sn= 1×2^2+2×2^3+.+（n-1)×2^n+n×2^(n+1) (2)
(2)-(1)：Sn=-2-2^2-2^3-.-2^n+n×2^(n+1）
=-2（1-2^n)/(1-2)+n×2^(n+1)
=(n-1)×2^(n+1)+2
2.Sn=1+2×2+3×2^2+.+n×2^(n-1) (3)
2Sn= 1×2+2×2^2+...+(n-1)×2^(n-1)+n×2^n (4)
(4)-(3)：Sn=-1-2-2^2-.-2^(n-1)+n×2^n
=-(1-2^n)/(1-2)+n×2^n
=(n-1)×2^n+1
3.Sn=1×(1/2)+2×(1/2)^2+3×(1/2)^3+.+n×(1/2)^n (5)
1/2Sn= 1×(1/2)^2+2×(1/2)^3+.+(n-1)×(1/2)^n+n×(1/2)^(n+1) (6)
(5)-(6)：1/2Sn=1/2+(1/2)^2+(1/2)^3+.+(1/2)^n+n×(1/2)^(n+1)
=1+(n-2)/2×(1/2)^n
Sn=(n-2)×(1/2)^n+2
4.Sn=1×2+3×2^2+5×2^3+.+(2n-1)×2^n (7)
2Sn= 1×2^2+3×2^3+.+(2n-3)×2^n+(2n-1)×2^(n+1) (8)
(8)-(7)：Sn=-2-2×2^2-2×2^3-.-2×2^n+(2n-1)×2^(n+1)
=-2-8[1-2^(n-1)]/(1-2)+(2n-1)×2^(n+1)
=(2n-3)×2^(n+1)+6

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)
两式相减有：-Sn=2^1+2^2+2^3......2^n-n*2^(n+1)
=2(1-2^n)/(1-2)-n*2^(n+1)
...

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)
两式相减有：-Sn=2^1+2^2+2^3......2^n-n*2^(n+1)
后面等比数列楼主会求了撒，后面的大同小异