Question

$\frac{1}{1*2}+ \frac{1}{2*3}+ \frac{1}{3*4}+...+ \frac{1}{99*100}$

Answer 1

<exp din cerinta>=1-1/2+1/2-1/3+1/3.......+1/99-1/100=1-1/100=99/100
reg pe care am fol e $\frac{1}{n(n+1)} = \frac{1}{n}- \frac{1}{n+1}$