题目

an=(2n-1)*2^n(n属于正整数),求Sn.注：

提问时间：2021-01-08

答案

Sn=a(1)+a(2)+.+a(n)=1*2^1+3*2^2.+(2n-1)*2^n
2Sn=2a(1)+2a(2)...+2a(n)
=1*2^2+3*2^3.+(2n-1)*2^(n+1)
Sn=2Sn-Sn=1*2^2+3*2^3.+(2n-1)*2^(n+1)-(1*2^1+3*2^2.+(2n-1)*2^n)
=(2n-1)*2^(n+1)-1*2^1+(1*2^2-3*2^2+3*2^3-5*2^3...+(2n-3)*2^n-(2n-1)*2^n)
=(2n-1)*2^(n+1)-2-(2^3+2^4...+2^(n+1))
=(2n-1)*2^(n+1)-(2^(n+2)-6)
(这一步利用了2+2+2^2+2^3...+2^(n+1)=2^(n+2))
=2^(n+1)*(2n-3)+6