Question

Найти сумму последовательности 1/6 + 1/12 + ... + 1/((n + 1) (n + 2))

Answer 1

$\displaystyle \dfrac{1}{2\cdot 3}+ \frac{1}{3\cdot 4}+...+ \frac{1}{(n+1)(n+2)} =\\ \\ \\ = \frac{3-2}{2\cdot 3} + \frac{4-3}{3\cdot 4} +...+ \frac{n+2-(n+1)}{(n+1)(n+2)} =\\ \\ \\ = \frac{1}{2}-\frac{1}{3} +\frac{1}{3}-\frac{1}{4} +...+ \frac{1}{n+1} -\frac{1}{n+2} =\frac{1}2}- \frac{1}{n+2}$