mateforum.ro

Se considera numerele : \( A=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{n}{n+1} \) si B\( =1\frac{1}{2}+2\frac{1}{3}+3\frac{1}{4}+n\frac{1}{n+1} \) , unde \( n\in\mathb{N}^* \) . Aratati ca numarul \( A+B-2n \) este natural .

\( A+B-2n=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.....+\frac{n}{n+1}+1\frac{1}{2} +\frac{1}{3}+3\frac{1}{4}+.....+n\frac{1}{n+1}-2n \)\( =\frac{1}{2}+\frac{1}{2}+\frac{2}{3}+\frac{1}{3}+\frac{3}{4}+\frac{1}{4}+.......+\frac{n}{n+1}+
+\frac{1}{n+1}+1+2+3+.....+n-2n \)\( =\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+.....+\frac{n+1}{n+1}+1+2+3+....+n- \)
\( -2n=n+\frac{n(n+1)}{2}-2n \)\( =\frac{2n+n(n+1)-4n}{2} \)\( =\frac{n(2+n+1-4)}{2} \)\( =\frac{n(n-1)}{2}\in \mathb{N}* \), deoarece produsul a oricare doua numere consecutive este par.