Question

数列an的前N项和为Sn=n²+3n+1,则通项公式是

Answer 1

当n=1时,

a1=S1=5

当n≥2时,

an=Sn-S(n-1)=[n²+3n+1]-[(n-1)²+3(n-1)+1]

=[n²-(n-1)²]+3[n-(n-1)]

=2n-1+3=2n+2

所以通项必须分段：

an={ 5 (n=1)

{ 2n+2 (n≥2)