各位数学大大，我想问问： 1. 1+1/(1+2)+1/(1+2+3)+......+1/(1+2+3+.....+99+100) = ? 2. 1/(1x2)+1/(2x3)+1/(3x4).....+1/(99x100)=? 备注:1/2=二分之一 ,请答我答案和计算方法.?

Sol
1
1+1/(1+2)+1/(1+2+3)+......+1/(1+2+3+.....+99+100)
=Σ(k=1 to 100)_1/(1+2+3+…+k)
=Σ(k=1 to 100)_2/[k(k+1)]
=2Σ(k=1 to 100)_1/[k(k+1)]
=2Σ(k=1 to 100)_[(k+1)－k]/[k(k+1)]
=2Σ(k=1 to 100)_{(k+1)/[k(k+1)]－k/[k(k+1)]}
=2Σ(k=1 to 100)_[1/k－1/(k+1)]
=2*[(1/1－1/2)+(1/2－1/3)+(1/3－1/4)+…+(1/100－1/101)]
=2*(1－1/101)
=2*100/101
=200/101
2. 1/(1x2)+1/(2x3)+1/(3x4).....+1/(99x100)
=Σ(k=1 to 99)_1/[k*(k+1)]
=Σ(k=1 to 99)_[(k+1)－k]/[k*(k+1)]
=Σ(k=1 to 99)_｛[(k+1)/[k*(k+1)]－k]/[k*(k+1)]｝
=Σ(k=1 to 99)_[1/k－1/(k+1)]
=(1－1/2)+(1/2－1/3)+(1/3－1/4)+…+(1/99－1/100)
=1－1/100
=99/100