设数列{an}的前n项和为Sn,若对任意正整数,都有Sn=n(a1+an)/2,证明{an}是等差数列.

an=Sn-Sn-1=n(a1+an)/2-(n-1)(a1+an-1)/2
2an=na1+nan-na1-nan-1+a1+an-1
(n-2)an=(n-1)*(an-1)-a1 (1)
同理
(n-1)*(an+1)=nan-a1 (2)
(1)-(2)
得到
(2n-2)an=(n-1)*(an-1)+(n-1)(an+1)
2an=an-1+an+1
所以an+1-an=an-an-1
得证