Question

已知数列{an}的前n项和Sn=n^2+n-1，则数列{an}的通项公式为

过程谢谢，貌似要分类

Answer 1

a1=S1=1^2+1-1=1
Sn=n^2+n-1
S(n-1)=(n-1)^2+(n-1)-1
=n^2-2n+1+n-2
=n^2-n-1

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

an=2n(n>=2)

Answer 2

n=1,an=1
n>1,an=2n