Question

Va rog daca puteti cat mai repede...

Answer 1

Avem $\displaystyle\frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3}\left(\displaystyle\frac{1}{k(k+1)(k+2)}-\displaystyle\frac{1}{(k+1)(k+2)(k+3)}\right)$
Dând lui k valori de la 1la 10, suma este egală cu
$\frac{1}{3}\left(\frac{1}{1\cdot 2\cdot 3}-\frac{1}{2\cdot 3\cdot 4}+\frac{1}{2\cdot 3\cdot 4}-\frac{1}{3\cdot 4\cdot 5}+\ldots +\frac{1}{10\cdot 11\cdot 12}-\frac{1}{11\cdot 12\cdot 13}\right)=\frac{1}{3}\left(\frac{1}{1\cdot 2\cdot 3}-\frac{1}{11\cdot 12\cdot 13}\right)$
Mai departe se pot face calculele.