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若n为正整数,求1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/(n+99)(n+100)的值
观察算式1/1X2=1-1/2 1/1x2+1/2x3=1-1/2+1/2-1/3=2/3.
人气:122 ℃ 时间:2020-09-30 23:08:33
解答
1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/(n+99)(n+100)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/(n+99)-1/(n+100)
=1/n-1/(n+100)
=100/n(n+100)
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