用数学归纳法证明1/(1x3)+1/(3x5)+1/(5x7)…1/(2n-1)(2n+1)=n/(2n+1)我证明完n=k+1后与结论不符,不知哪错了当n=k时成立即1/(1x3)+1/(3x5)+1/(5x7)…1/(2k-1)(2k+1)=k/(2k+1)则n=k+1时1/(1x3)+1/(3x5)+1/(5x7)…1/(2k-1)(2k+1)+1/(2k+1
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问题描述：
用数学归纳法证明1/(1x3)+1/(3x5)+1/(5x7)…1/(2n-1)(2n+1)=n/(2n+1)
我证明完n=k+1后与结论不符,不知哪错了
当n=k时成立
即1/(1x3)+1/(3x5)+1/(5x7)…1/(2k-1)(2k+1)=k/(2k+1)
则n=k+1时
1/(1x3)+1/(3x5)+1/(5x7)…1/(2k-1)(2k+1)+1/(2k+1)(2k+3)
=k/(2k+1)+1/(2k+1)(2k+3)
=2k^2+3k+1/(2k+1)(2k+3)
=(k+1/2)(k+1)/(2k+1)(2k+3)
=(k+1)/2(2k+3)
而原式应为(k+1)/(2k+3)
陆绍飞回答：
=k/(2k+1)+1/(2k+1)(2k+3)=（2k+1）（k+1/（2k+3））=（2k+1）（(2k方+3k+1)/（2k+3））=1/（2k+1）*（(2k+1)（k+1）/（2k+3））=（k+1）/（2k+3）成立你算的=(2k^2+3k+1）/(2k+1)(2k+3)=（2k+1）（k+1）/(2k+1)(2k+...
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