1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+100)
= 2/(1×2)+2/(2×3)+2/(3×4)+……+2/(100×101)
= 2×[1/(1×2)+1/(2×3)+1/(3×4)+……+1/(100×101)]
= 2×[(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
= 2×(1-1/101)
= 2×(100/101)
= 200/101
= 1又99/101
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