题目

等差数列An=2n+1 Bn=1/An^2-1 求Bn前N项和Tn

提问时间：2020-11-02

答案

Bn=1/An^2-1=1/[(2n+1)^2-1]=1/(4n^2+4n)=1/4*1/[n*(n+1)]
又∵1/[n*(n+1)]=1/n-1/(n+1)]
∴Tn=1/4*[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=1/4*[1/1-1/(n+1)]
=1/4*[(n+1)/(n+1)-1/(n+1)]
=1/4*[(n+1-1)/(n+1)]
=n/(4n+4)
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