数列an的前n项和sn=n²+3n,求an的通项公式
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![数列an的前n项和sn=n²+3n,求an的通项公式](/uploads/image/z/1240678-46-8.jpg?t=%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%3Dn%26%23178%3B%2B3n%2C%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
数列an的前n项和sn=n²+3n,求an的通项公式
数列an的前n项和sn=n²+3n,求an的通项公式
数列an的前n项和sn=n²+3n,求an的通项公式
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
Sn= n^2+3n (1)
S(n-1) = (n-1)^2+3(n-1) (2)
(1)-(2)
an= 2n+2
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