数列{an}的通项公式an=log2(n+1)-log2(n+2),设{an}的前n项和为Sn,则使Sn

问题描述:

数列{an}的通项公式an=log2(n+1)-log2(n+2),设{an}的前n项和为Sn,则使Sn

Sn=[-log2(n+2)+log2(n+1)]+[-log2(n+1)+log2(n)]+[-log2(n)+log2(n-1)]+...+[-log2(2)+log2(1)]=-log2(n+2)+log2(1)=-log2(n+2)5=log2(2^5)n+2>2^5=32n>30正整数n的最小值为31