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一道高一数列求和 (2n+1)/[n^2*(n+1)^2]
这是通项 求Sn
人气:496 ℃ 时间:2020-02-06 00:47:16
解答
An=(2n+1)/[n^2*(n+1)^2]
=[1/(n^2)]-[1/(n+1)^2]
A1=(1/1)-[1/(2^2)]
A2=[1/(2^2)-[1/(3^2)]
A3=[1/(3^2)-[1/(4^2)]
……
An-1=[1/(n-1)^2]-[1/n^2]
Sn=A1+A2+A3+A4+……+An-1+An
=(1/1)-[1/(2^2)]+[1/(2^2)-[1/(3^2)]+[1/(3^2)-[1/(4^2)]+……+[1/(n-1)^2]-[1/n^2]+[1/(n^2)]-[1/(n+1)^2]
=1-[1/(n+1)^2]
=(n^2+2n)/[(n+1)^2]
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