Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$ 
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.

I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see how that helps.

Note that we can write
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(\frac{(m+2)-m}{m(m+1)(m+2)}\right)\\\\
&=\frac12\left(\frac{1}{m(m+1)}-\frac{1}{(m+1)(m+2)}\right)
\end{align}$$
So, letting $a_m=\frac{1}{m(m+1)}$, we see that
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(a_m-a_{m+1}\right)
\end{align}$$
from which we find
$$\begin{align}
\sum_{m=1}^n \frac{1}{m(m+1)(m+2)}&=\sum_{m=1}^n \frac12\left(a_m-a_{m+1}\right)\\\\
&=\frac12(a_1-a_{n+1})\\\\
&=\frac14-\frac12 \frac{1}{(n+1)(n+2)}
\end{align}$$
as was to be shown!
