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1+(1/1+2)+(1/1+2+3)+...+(1/1+2+3+...100)=?
人气:448 ℃ 时间:2020-04-05 22:56:41
解答
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...100)
=1+1/3+1/6+1/10+...+1/5050
=2/2+2/6+2/12+2/20+...+2/10100
=2/(1×2)+2/(2×3)+2/(3×4)+2/(4×5)+...+2/(100×101)
=2(1/1-1/2)+2(1/2-1/3)+2(1/3-1/4)+2(1/4-1/5)+...+2(1/100-1/101)
=2(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/100-1/101)
=2(1-1/101)
=200/101
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