数列{an}的前n项和sn=n^2-n,则a4

题目

数列{an}的前n项和sn=n^2-n,则a4

答案

Sn=n^2-n
S(n-1)=(n-1)^2-(n-1)
=n^2-2n+1-n+1
=n^2-3n+2
Sn-S(n-1)=An
=n^2-n-n^2+3n-2
=2n-2
a4=2*4-2=6