Question

1 supra 1x2+1s upra2x3 +.....+1 supra n(n+1)=n supra n+1

Answer 1

$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}+...+ \frac{1}{n(n+1)}$

$\frac{1}{1*2} + \frac{1}{2*3} = \frac{3}{2*3} + \frac{1}{2*3} = \frac{4}{6} = \frac{2}{3}$
 observam ca daca n=2, atunci suma = $\frac{2}{2+1}$

$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}= \frac{2}{3} + \frac{1}{3*4}= \frac{2*4}{3*4}+ \frac{1}{3*4}= \frac{9}{12} = \frac{3}{4}$

 observam ca daca n=3, atunci suma = $\frac{3}{3+1}$

=> adaugand termen dupa termen, suma va fi : $\frac{n}{n+1}$
=>
$\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}+...+ \frac{1}{n(n+1)} = \frac{n}{n+1}$