数列{(2n-1)⼀2^n}的前n项和Sn

简单计算一下即可，答案如图所示

分析

Sn=（2-1）/2^1+（2×2-1）/2^2+.+（2n-3）/2^（n-1）+（2n-1）/2^n①
2Sn=（2-1）+（2×2-1）/2^1+（2×3-1）/2^3+.+（2n-1）/2^（n-1）②
②减①=Sn=1+2（1/2^1+1/2^2+.+1/2^（n-1））-（2n-1）/2^n
=1+2（1-1/2^（n-1））-（2n-1）/2ⁿ
=3-4/2ⁿ-（2n-1）/2ⁿ
=3-（2n+3）/2ⁿ
综上,{an}的前n项和Sn=3-（2n+3）/2ⁿ.

let
S = 1.(1/2)^0 +2.(1/2)^1+....+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1 +2.(1/2)^2+....+n.(1/2)^n (2)
(1)-(2)
(1/2)S = [1+1/2+..+1/2^(n-1) ] - n.(1/2)^n
= 2(1- (1/2)^n ) - n.(1/2)^n
S =4(1- (1/2)^n ) - 2n.(1/2)^n
an = (2n-1)/2^n
= n .(1/2)^(n-1) - 1/2^n
Sn = a1+a2+...+an
= S - (1 - 1/2^n)
=4(1- (1/2)^n ) - 2n.(1/2)^n -(1 - 1/2^n)
=3(1- (1/2)^n ) - 2n.(1/2)^n
=3 -(2n-3) .(1/2)^n

Sn=-(2n+3)·(1/2)^n+3