题目

正项数列{an}的前n项和Sn满足:Sn^2-(n^2+n-1)Sn-(n^2+n)=0
求数列的通项公式an
令bn=n+1/（n+2）^2*an^2,数列的前n项和为Tn,证明对任意的数,都有Tn<5/64

提问时间：2021-01-08

答案

Sn^2-(n^2+n-1)Sn-(n^2+n)=0
[Sn+1][Sn-(n^2+n)]=0
∵an>0
∴Sn+1≠0
∴Sn=n^2+n
a1=S1=2
n≥2时,an=Sn-S(n-1)=2n
∴{an}的通项公式为an=2n
bn=（n+1)/[(n+2)^2(an)^2]
=(n+1)/[4n^2(n+2)^2]
=1/16[1/n^2-1/(n+2)^2]
Tn=1/16[1-1/9+1/4-1/16+1/9-1/25+.+1/(n-1)^2-1/(n+1)^2+1/n^2-1/(n+2)^2]
=1/16[1+1/4-1/(n+1)^2-1/(n+2)^2]
=1/16*[5/4-1/(n+1)^2-1/(n+2)^2]