How to proceed to show this hold by induction? 
Show that $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{n+2}{n(n+1)(n+3)}=\frac16\left[\frac{29}6-\frac4{n+1}-\frac1{n+2}-\frac1{n+3}\right],\text{ for $n\in\mathbb N$}.$$

I try induction. For $n=1$, it is trivial and then let it is true for some $k\in\mathbb N.$ Then for $k+1$,we have to show $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{k+3}{(k+1)(k+2)(k+4)}=\frac16\left[\frac{29}6-\frac4{k+2}-\frac1{k+3}-\frac1{k+4}\right],$$ which is quivalent to show  $$\frac16\left[\frac{29}6-\frac4{k+1}-\frac1{k+2}-\frac1{k+3}\right]+\frac{k+3}{(k+1)(k+2)(k+4)}=\frac16\left[\frac{29}6-\frac4{k+2}-\frac1{k+3}-\frac1{k+4}\right]$$ by induction hypothesis. Now is there any easier trick other than long and tedious computation?
 A: $$\begin{eqnarray*}\frac{n+2}{n(n+1)(n+3)}&=&\frac{1}{n(n+1)}\left(1-\frac{1}{n+3}\right)\\&=&\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(1-\frac{1}{n+3}\right)\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n(n+3)}+\frac{1}{(n+1)(n+3)}\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\\&=&\frac{\frac{2}{3}}{n}-\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{6}}{n+3}\end{eqnarray*}$$
(that can be done through the residue theorem, too), hence, by setting $H_n=\sum_{k=1}^{n}\frac{1}{k}$:
$$\begin{eqnarray*} \sum_{n=1}^{N}\frac{n+2}{n(n+1)(n+3)}&=&\frac{2}{3}\sum_{n=1}^{N}\frac{1}{n}-\frac{1}{2}\sum_{n=1}^{N}\frac{1}{n+1}-\frac{1}{6}\sum_{n=1}^{N}\frac{1}{n+3}\\&=&\frac{2}{3}\sum_{n=1}^{N}\frac{1}{n}-\frac{1}{2}\sum_{n=2}^{N+1}\frac{1}{n}-\frac{1}{6}\sum_{n=4}^{N+3}\frac{1}{n}\\&=&\left(\frac{2}{3}-\frac{1}{2}-\frac{1}{6}\right)H_N-\frac{1}{2}\left(\frac{1}{N+1}-H_1\right)-\frac{1}{6}\left(\frac{1}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}-H_3\right)\\&=&\frac{H_1}{2}+\frac{H_3}{6}-\frac{1}{6}\left(\frac{4}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}\right)\\&=&\color{red}{\frac{29}{36}-\frac{1}{6}\left(\frac{4}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}\right)} \end{eqnarray*}$$
as wanted.
