Question

已知数列{an}的前n项和为sn=n²+c(c为常数),求数列{an}的通项公式,并判断{an}是不是等差数列

Answer 1

Sn=n^2+c

Sn-1=(n-1)^2+c

an=Sn-Sn-1

=n^2-(n-1)^2

=n^2-n^2+2n-1

=2n-1

an=2n-1

an-a(n-1)=2n-1-2(n-1)+1=2,为定值.

a1=2-1=1

数列为首项是1,公差是2的等差数列.