Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.
Do I have to use something else? Or am I missing some easy manipulation?

\begin{align*}
T_r
&= \frac{1}{r(r+1)} - \frac{1}{r(r+1)(r+3)} \\
&= \frac{1}{r(r+1)}- \left(\frac{1}{2}\cdot \frac{(r+3)-(r+1)}{r(r+1)(r+3)}\right) \\
&= \frac{1}{r}-\frac{1}{r+1}- \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot \frac{1}{r(r+3)} \\
&= \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot\frac{1}{3}\left(\frac{1}{r}-\frac{1}{r+3}\right) \\
&= \frac{2}{3}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{6}\left(\frac{1}{r+1}-\frac{1}{r+2}\right)+ \frac{1}{6}\left(\frac{1}{r+2}-\frac{1}{r+3}\right)
\end{align*}
You can now see telescope and quicky add the terms.
