已知数列{an}的通项公式an=log2n+1分之n+2,设其前n项为S 求Sn?

题目

已知数列{an}的通项公式an=log2n+1分之n+2,设其前n项为S 求Sn?

答案

因an=log2[(n+2)/(n+1)]=log2(n+2)-log2(n+1),n应该从1开始.
所以Sn=log2(3)-log2(2)+log2(4)-log2(3)+...+log2(n+2)-log2(n+1)
=log2(n+2)-log2(2)
=log2(n+2)-1