(1/1x3)+(1/3x5)+(1/5x7)+……+(1/49x51)=如题

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(1/1x3)+(1/3x5)+(1/5x7)+……+(1/49x51)=如题
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(1/1x3)+(1/3x5)+(1/5x7)+……+(1/49x51)=如题
(1/1x3)+(1/3x5)+(1/5x7)+……+(1/49x51)=
如题

(1/1x3)+(1/3x5)+(1/5x7)+……+(1/49x51)=如题
设数列An通项=1/[(2n-1)(2n+1)] n≥1且n∈Z
An= 1/[(2n-1)(2n+1)]= 1/2* [1/(2n-1) -1/(2n+1)]
记数列An的前n项和为Sn.
则:Sn=1/2[1-1/(2n+1)],(累加相消)
∴ Sn=(1/1x3)+(1/3x5)+(1/5x7)+……+(1/49x51)=1/2[1-1/(2*25+1)]=25/51