1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)=17/35,求n的值

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1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)=17/35,求n的值1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)=17/3

1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)=17/35,求n的值
1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)=17/35,求n的值

1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)=17/35,求n的值
1/(|x3)+1/(3x5)+|/(5x7)+,..+1/(2n-1)(2n+1)
=[1/2]x[1/1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1)]
=[1/2]x[1-1/(2n+1)]
=n/(2n+1)
=17/35
=>
n=17

首先把每一个分式拆成两项之差,即1/(1x3)=1/2(1/1-1/3),,1/(3x5)=1/2(1/3-1/5),
1/(5x7)=1/2(1/5-1/7)......1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]
将他们相加并都提1/2,则原式= (1/2)x[ (1-1/3)+(1/3-1/5)+(1/5-1/7)+……+1/(2n-1)...

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首先把每一个分式拆成两项之差,即1/(1x3)=1/2(1/1-1/3),,1/(3x5)=1/2(1/3-1/5),
1/(5x7)=1/2(1/5-1/7)......1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]
将他们相加并都提1/2,则原式= (1/2)x[ (1-1/3)+(1/3-1/5)+(1/5-1/7)+……+1/(2n-1)-1/(2n+1)]
=(1/2)x[ (1- 1/(2n+1)]=n/(2n+1)=17/35
n=17

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