数列an的前n项和sn=n²+3n,求an的通项公式
问题描述：
数列an的前n项和sn=n²+3n,求an的通项公式
答
n=1,a1=S1=1+3=4
n>=2,an=Sn-S(n-1)=n^2+3n-(n-1)^2-3(n-1)=2n-1+3=2n+2
a1=2+2=4,符合
故有an=2n+2
答
Sn= n^2+3n (1)
S(n-1) = (n-1)^2+3(n-1) (2)
(1)-(2)
an= 2n+2
答
Sn=n^2+3n
S(n-1)=(n-1)^2+3(n-1)
=n^2-2n+1+3n-3
=n^2+3n+(-2n-2)
Sn-S(n-1)
=an
=n^2+3n-n^2-3n+2n+2
=2n+2