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数学
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1+(1+2)/1+(1+2+3)/1+...+(1+2+3+...+100)/1等于多少?
人气:322 ℃ 时间:2020-04-13 10:35:38
解答
∵1+2+3+...+n = n*(n+1)/2
∴1/(1+2+3+...+n) = 2/n*(n+1) =2*[1/n - 1/(n+1)]
从而原式=1+2*(1/2-1/3+1/3-1/4+...+1/100-1/101)
=1+2*(1/2-1/101)=200/101
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