已知数列{an}其通项公式为an=2的n次方分之2n-1 求数列的前n项和 Sn

Question

如题所述

Answer 1

解答：
错位相减求和
Sn=1/2^1+3/2^2+5/2^3+.......+(2n-3)/2^(n-1)+(2n-1)/2^n ①‘
①×1/2
(1/2)Sn= 1/2^2+3/2^3+...............................+(2n-3)/2^n+(2n-1)/2^(n+1) ②
①-②
(1/2)Sn=1/2+ 1/2 +1/2^2+.................................+1/2^(n-1)-(2n-1)/2^(n+1)
=1/2+ 1-1/2^(n-1)-(2n-1)/2^(n+1)
Sn =1 + 2 - 4/2^n-(2n-1)/2^n
∴ Sn=3-(2n+3)/2^n