Question

$\frac{3}{6*9} + \frac{3}{9*12} + \frac{3}{12*15} +...... + \frac{3}{21*24} =$

Answer 1

      
$\frac{3}{6*9} + \frac{3}{9*12} + \frac{3}{12*15} +...... + \frac{3}{21*24} = \\ \text{Dam factor comun pe } \frac{3}{3*3} \\ =\frac{3}{3*3}( \frac{1}{2*3}+ \frac{1}{3*4}+ \frac{1}{4*5}+.......+ \frac{1}{7*8})= \\ \text{Stim ca: } \frac{1}{n*(n+1)}= \frac{1}{n}- \frac{1}{n+1} \\ = \frac{3}{3*3}(\frac{1}{2}-\frac{1}{3}+ \frac{1}{3}-\frac{1}{4}+ \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+ \frac{1}{7}-\frac{1}{8})= \\ =\frac{3}{3*3}(\frac{1}{2}-\frac{1}{8})=$

$=\frac{1}{3}*\frac{4-1}{8}= \\ =\frac{1}{3}*\frac{3 }{8}= \frac{1}{8}$