116问答网 > 已知数列的前n项和Sn=n^2+1⼀2n，求an并证明｛an｝是等差数列

已知数列的前n项和Sn=n^2+1⼀2n，求an并证明｛an｝是等差数列

2025-06-25 21:01:38

推荐回答（1个）

回答1：

a1=S1=1^2+1/2*1=3/2

Sn=n^2+n/2
S(n-1)=(n-1)^2+(n-1)/2
=n^2-2n+1+n/2-1/2
=n^2-3n/2+1/2

an=Sn-S(n-1)
=n^2+n/2-(n^2-3n/2+1/2)
=n^2+n/2-n^2+3n/2-1/2
=2n-1/2

a(n-1)
=2(n-1)-1/2
=2n-2-1/2
=2n-5/2

an-a(n-1)
=2n-1/2-(2n-5/2)
=2n-1/2-2n+5/2
=2

所以an是以a1=3/2，公差为2的等差数列