Sn=1*2+3*2^2+5*2^3+……+(2n-1)*2^n 求Sn=

Sn=1*2+3*2^2+5*2^3+……+(2n-1)*2^n (1)
2sn= 1*2^2+3*2^3+……+(2n-3))*2^n+(2n-1)*2^(n+1) (2)
(1)-(2)
-sn=1*2+2^3+2^4+……+2^(n+1)+(2n-1)*2^(n+1)
-sn=1*2-2^(n+2)+(2n-1)*2^(n+1)
sn=2^(n+2)-(2n-1)*2^(n+1)-2

Sn=1*2+3*2^2+5*2^3+……+(2n-1)*2^n
(1/2)Sn=1+3*2+5*2^2+.....+(2n-1)*2^(n-1)
(1/2)Sn-Sn=1+2[2+2^2+...+2^(n-1)]-(2n-1)*2^n
-(1/2)Sn=1+2*2[2^(n-1)-1]/(2-1)-(2n-1)*2^n
=1+2*2^n-4-(2n-1)*2^n
=-3-(2n-3)*2^n
所以Sn=6+(2n-3)*2^(n+1)

使用错位相减法：
Sn=1*2+3*2^2+5*2^3+……+(2n-1)*2^n
2Sn=1*2^2+3*2^3+...+(2n-1)*2^(n+1)
两式相减得：-Sn=1*2+2*2^2+2*2^3+..+2*2^n—(2n-1)*2^(n+1)
=1*2+[2*2^2+2*2^3+..+2*2^n]—(2n-1)*2^(n+1)

所以sn=n*2^(2 + n) )- 3*2^（n+1）+6

Sn=1*2+(2+1)*2^2+(4+1)*2^3+......+[2*(n-1)+1]*2^n

Sn=1*2+3*2^2+5*2^3+……+(2n-1)*2^n2Sn=1*2^2+3*2^3+...+(2n-1)*2^(n+1)相减得-Sn=1*2+2*2^2+2*2^3+..+2*2^n-(2n-1)*2^(n+1)=1*2+[2*2^2+2*2^3+..+2*2^n]-(2n-1)*2^(n+1)之后[]里的用等比数列求和公式算化简得sn=2 ...