Question

已知数列{an}的前n项和为Sn=n^2+2n,求数列{an}的通项公式

Answer 1

Sn=n^2+2n
S(n-1)=(n-1)^2+2(n-1)
=n^2-2n+1+2n-2
=n^2-1
an=Sn-S(n-1)
=n^2+2n-(n^2-1)
=2n+1