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

错位相减求和
Sn=1/2^1+3/2^2+5/2^3+.+(2n-3)/2^(n-1)+(2n-1)/2^n①‘
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