1/1·3+1/2·4+1/3·5+```````+1/n(n+2)等于多少?

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1/1·3+1/2·4+1/3·5+```````+1/n(n+2)等于多少?
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1/1·3+1/2·4+1/3·5+```````+1/n(n+2)等于多少?
1/1·3+1/2·4+1/3·5+```````+1/n(n+2)等于多少?

1/1·3+1/2·4+1/3·5+```````+1/n(n+2)等于多少?
1/n(n+2)=[1/n-1/(n+2)]/2
就是1/1*3=(1/1-1/3)/2
1/2*4=(1/2-1/4)/2
1/3*5=(1/3-1/5)/2
这样上面的式子有能约掉的项
剩下的＝[1/1+1/2-1/(n+1)-1/(n+2)]/2
然后再化简下就得结果了

1/n(n+2)=[1/n-1/(n+2)]/2
1/1·3+1/2·4+1/3·5+```````+1/n(n+2)
=1/2*[1-1/3+1/2-1/4+1/3-1/5+…………+1/n-1/(n+2)]
=1/2*[1/3-1/(n+1)-1/(n+2)]
=(n^2+n-1)/[6(n+1)(n+2)]

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