Question

已知数列{an}的前n项和Sn=2n^2+n+1，求该数列的通项公式

Answer 1

Sn－Sn-1=an=2n²+n+1-2(n-1)²-(n-1)-1
=4n-1
即an=4n-1

Answer 2

An=Sn-S(n-1)=2n^2+n+1-[2(n-1)^2+n-1+1]
=2n^2+n+1-(2n^2-4n+2+n)
=4n-1