Question

已知数列{an}的前n项和sn=n²+2n,则an=

Answer 1

n=1时,a1=S1=1+2=3

n≥2时,

Sn=n²+2n S(n-1)=(n-1)²+2(n-1)

an=Sn-S(n-1)=n²+2n-(n-1)²-2(n-1)=2n+1

n=1时,a1=2+1=3,同样满足.

数列{an}的通项公式为an=2n+1.