题目

数列(2n-3)/2^(n-3)的前n项和为?

提问时间：2020-11-11

答案

设数列(2n-3/2^(n-3))的前n项和为Sn 则Sn=-4+2+3+5/2+7/2^2+9/2^3+.+(2n-5)/2^(n-4)+(2n-3)/2^(n-3) =1+5/2+7/2^2+9/2^3+.+(2n-5)/2^(n-4)+(2n-3)/2^(n-3).(1) 把(1)式两端同乘以(1/2),得 Sn/2=1/2+5/2^2+7/2^3+9/2^4+.+(2n-5)/2^(n-3)+(2n-3)/2^(n-2).(2) (1)-(2),得 Sn/2=1/2+5/2+2/2^2+2/2^3+.+2/2^(n-3)-(2n-3)/2^(n-2) =2+1+1/2+1/2^2+.+1/2^(n-4)-(2n-3)/2^(n-2) =2+(1-1/2^(n-3))/(1-1/2)-(2n-3)/2^(n-2) =2+2-1/2^(n-4)-(2n-3)/2^(n-2) =4-(2n+1)/2^(n-2) 故Sn=8-(2n+1)/2^(n-3).