题目

已知公差为d的等差数列的前n项之和为Sn,则Sn+1/n+1-Sn/n

提问时间：2021-01-05

答案

Sn+1/n+1=[(n+1)a1+n*(n+1)d/2]/(n+1)=a1+n*d/2
Sn/n=[n*a1+n*(n-1)d/2]/n=a1+n*d/2-d/2
Sn+1/n+1-Sn/n =(a1+n*d/2)-(a1+n*d/2-d/2)
=d/2