1+(1+2)/1+(1+2+3)/1+...+(1+2+3+...+100)/1等于多少?
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1+(1+2)/1+(1+2+3)/1+...+(1+2+3+...+100)/1等于多少?
![1+(1+2)/1+(1+2+3)/1+...+(1+2+3+...+100)/1等于多少?](/uploads/image/z/4042757-29-7.jpg?t=1%2B%281%2B2%29%2F1%2B%281%2B2%2B3%29%2F1%2B...%2B%281%2B2%2B3%2B...%2B100%29%2F1%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91%3F)
∵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
∴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