Question

Geniu/Moderator(Dau coroana pentru un raspuns corect si o explicatie buna!)

Calculati:

$\frac{1}{1*2*3}+\frac{1}{2*3*4}+\frac{1}{3*4*5}+...+\frac{1}{n(n+1)(n+2)}$

Answer 1

$\frac{1}{(k-1)k(k+1)}= \frac{1}{k} \cdot \frac{1}{(k-1)(k+1)}= \frac{1}{k} \cdot \frac{1}{2} ( \frac{1}{k-1}- \frac{1}{k+1})= \frac{1}{2} \Big( \frac{1}{(k-1)k}- \frac{1}{k(k+1)} \Big) . \\ \\ S= \frac{1}{2} \Big( \frac{1}{1 \cdot 2}- \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4}+ \frac{1}{3 \cdot 4}- \frac{1}{4 \cdot 5} +...+ \frac{1}{n(n+1)}- \frac{1}{(n+1)(n+2)} \Big)= \\ \\ ~~~= \frac{1}{2} \Big( \frac{1}{1 \cdot 2} - \frac{1}{(n+1)(n+2)} \Big)=$
$.~= \frac{1}{2} \Big( 1-\frac{1}{2} - \frac{1}{n+1}+ \frac{1}{n+2} \Big)= \\ \\ ~~~= \frac{1}{2} \Big( \frac{1}{2}- \frac{n+2}{(n+1)(n+2)}+ \frac{n+1}{(n+1)(n+2)} \Big)= \\ \\~~~= \frac{1}{2} \Big( \frac{1}{2}- \frac{1}{(n+1)(n+2)} \Big)$

$IDENTITATE:~ \frac{1}{k(k+m)}= \frac{1}{m} \cdot \frac{k+m-k}{k(k+m)}= \frac{1}{m} \cdot \Big( \frac{k+m}{k(k+m)}- \frac{k}{k(k+m)} \Big)= \\ \\ = \frac{1}{m} \Big( \frac{1}{k}- \frac{1}{k+m} \Big).$