[tex] \frac{2}{n+1}+ \frac{4}{n+1}+ \frac{6}{n+1}+.... \frac{2n}{n+1}= \\ \\ \frac{2}{n+1} (1+2+3+......+n)= \\ \\ \frac{2}{n+1} *[ \frac{n(n+1)}{2} ]= \\ \\ =n[/tex]
unde 1+2+3+...n- suma Gauss cu formula de calcul [ n(n+1)]:2
numerele sunt consecutive si incep din 1=suma Gauss
