设数列{an}的前n项和是Sn=2n^2+3n+2,求an的通项公式并指出数列是否为等差数列?

当n=1时,
a(1)=S(1)=2*1^2+3*1+2=7
当n>1时,
a(n)
=S(n)-S(n-1)
=2n^2+3n+2-2(n-1)^2-3(n-1)-2
=4n-2+3
=4n+1
a1=7显然不满足4n+1=5,所以不是等差数列