Unable to prove an assertion with induction I need to prove: 
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle \sum_{k=1}^{n+1}\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
which can be summarised to: 
$ \displaystyle \sum_{k=1}^{n}\frac{1}{(5k + 1) (5k + 6)} + \frac{1}{(5(n+1) + 1) (5(n+1) + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
Using the Induction Hyptohesesis this turns to:
$ ( \frac{1}{30} - \frac{1}{5(5n + 6)} ) + \frac{1}{(5n + 6 ) (5n+6+5)}  $ 
when we let s = 5n +6 we get:
$  \frac{1}{30} - \frac{1}{5s}  + \frac{1}{(s) (s+5)} = \frac{1}{30} - \frac{1}{5(s + 5)}  $ 
substracting both sides by $\frac{1}{30}$ yields: 
$- \frac{1}{5s}  + \frac{1}{(s) (s+5)} = - \frac{1}{5(s + 5)}$ and this is where I'm stuck. I'm working on this for more than 6h but my mathematical foundation is too weak to get a viable solution. Please help me.

Let $ n $ be a positive integer.
Observe that : $ \left(\forall k\in\mathbb{N}\right),\ \frac{1}{\left(5k+1\right)\left(5k+6\right)}=\frac{1}{5}\left(\frac{1}{5k+1}-\frac{1}{5k+6}\right) \cdot $
Thus, \begin{aligned} \sum\limits_{k=1}^{n}{\frac{1}{\left(5k+1\right)\left(5k+6\right)}}&=\frac{1}{5}\left(\sum\limits_{k=1}^{n}{\frac{1}{5k+1}}-\sum\limits_{k=1}^{n}{\frac{1}{5k+6}}\right)\\ &=\frac{1}{5}\left(\sum\limits_{k=0}^{n-1}{\frac{1}{5k+6}}-\sum\limits_{k=1}^{n}{\frac{1}{5k+6}}\right)\\ &=\frac{1}{5}\left(\frac{1}{6}-\frac{1}{5n+6}\right)\\ \sum\limits_{k=1}^{n}{\frac{1}{\left(5k+1\right)\left(5k+6\right)}}&=\frac{1}{30}-\frac{1}{5\left(5n+6\right)} \end{aligned}
