计算2/2²-1+2/4²-1+2/6²-1+2/8²-1+2/10²-1+2/12²-1.要有算式和答案

计算2/2²-1+2/4²-1+2/6²-1+2/8²-1+2/10²-1+2/12²-1.要有算式和答案
计算2/2²-1+2/4²-1+2/6²-1+2/8²-1+2/10²-1+2/12²-1.
要有算式和答案

计算2/2²-1+2/4²-1+2/6²-1+2/8²-1+2/10²-1+2/12²-1.要有算式和答案
2/(2²-1)+2/(4²-1)+2/(6²-1)+2/(8²-1)+2/(10²-1)+2/(12²-1)
=2/[(2-1)×(2+1)]+2/[(4-1)×(4+1)]+2/[(6-1)×(6+1)]+2/[(8-1)×(8+1)]+2/[(10-1)×(10+1)]+2/[(12-1)×(12+1)]
=2/(1×3)+2/(3×5)+2/(5×7)+2/(7×9)+2/(9×11)+2/(11×13)
=(1-1/3)+(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+(1/11-1/13)
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
=1-1/13
=12/13
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有新问题请求助,

您好：

2/2²-1+2/4²-1+2/6²-1+2/8²-1+2/10²-1+2/12²-1.
=2/(2+1)(2-1)+2/(4+1)(4-1)+2/(6+1)(6-1)+2/(8+1)(8-1)+2/(10+1)(10-1)+2/(12+1)(12-1)
=2/1x3+2/3x5+2/5x7+2...

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您好：

2/2²-1+2/4²-1+2/6²-1+2/8²-1+2/10²-1+2/12²-1.
=2/(2+1)(2-1)+2/(4+1)(4-1)+2/(6+1)(6-1)+2/(8+1)(8-1)+2/(10+1)(10-1)+2/(12+1)(12-1)
=2/1x3+2/3x5+2/5x7+2/7x9+2/9x11+2/11x13
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
=1-1/13
=12/13


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收起

可以看做数列求和，an=2/(2n)²-1，求Sn，其中n=6.
对an进行变形，an=2/(2n-1)(2n+1)=1/(2n-1)-1/(2n+1)
所以S6=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13=12/13