已知数列的前n项和Sn=n^2+1/2n,求an并证明{an}是等差数列

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/03 05:20:10
已知数列的前n项和Sn=n^2+1/2n,求an并证明{an}是等差数列
xQK 0NwM/ѽ(z BEq/Ƥ &ƵL2ofxy }~dzly{

已知数列的前n项和Sn=n^2+1/2n,求an并证明{an}是等差数列
已知数列的前n项和Sn=n^2+1/2n,求an并证明{an}是等差数列

已知数列的前n项和Sn=n^2+1/2n,求an并证明{an}是等差数列
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