Sn=1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+.+1/[(2n)^2-1]

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/11/07 17:57:16

Sn=1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+.+1/[(2n)^2-1]
求和

Sn=1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+.+1/[(2n)^2-1]
=1/(2+1)(2-1)+1/(4+1)(4-1)+1/(6+1)(6-1)+.+1/(2n+1)(2n-1)
=1/(3*1)+1/(5*3)+1/(7*5)+.+1/(2n+1)(2n-1)
因为1/(2n+1)(2n-1)=0.5*[1/(2n-1)-1/(2n+1)]
所以
Sn=0.5*[1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1)]
=0.5*[1-1/(2n +1 )]
=n/(2n+1)

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