Question

已知数列{an}为等差数列,它的前n项和为Sn,若Sn=(n+1)^2+λ,则λ的值是?

Answer 1

a1=s1=(1+1)^2+λ
a1=4+λ
Sn=(n+1)^2+λ
S(n-1)=n^2+λ
an=Sn-S(n-1)
=(n+1)^2+λ-n^2-λ
=n^2+2n+1+λ-n^2-λ
=2n+1
a1=2*1+1=2+1=3
3=4+λ
λ=-1