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.
 A: You started fine. Note that to prove by induction a sequence of equalities of the type$$a_1+a_2+\cdots+a_n=b_n$$is the same thing as proving that $a_1=b_1$ and that, for each $n\in\mathbb N$, $a_{n+1}=b_{n+1}-b_n$. So, you should prove that\begin{align}\frac1{5\bigl(5(n+1)+1\bigr)\bigl(5(n+1)+6\bigr)}&=\frac1{30}-\frac1{5\bigl(5(n+1)+6\bigr)}-\left(\frac1{30}-\frac1{5(5n+6)}\right)\\&=\frac1{5(5n+6)}-\frac1{5(5n+11)}.\end{align}But\begin{align}\frac1{5\bigl(5(n+1)+1\bigr)\bigl(5(n+1)+6\bigr)}&=\frac1{5(5n+6)(5n+11)}\\&=\frac1{5(5n+6)}-\frac1{5(5n+11)}.\end{align}
A: Your proof by induction is complete as soon as you compute the difference $$\frac{1}{5s} - \frac{1}{5(s + 5)}=\frac{(s+5)-s}{5s(s + 5)}=\frac{1}{s(s + 5)}.$$
There is also a direct proof: the sum is telescopic
$$ \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)}=\frac{1}{5}\sum_{k=1}^n\left(\frac{1}{5k + 1}-\frac{1}{5(k+1) + 1}\right)=\frac{1}{5}\sum_{k=1}^n\frac{1}{5k + 1}-\frac{1}{5}\sum_{k'=2}^{n+1}\frac{1}{5k' + 1}\\=\frac{1}{5}\sum_{k=1}^n\frac{1}{5\cdot 1 + 1}
-\frac{1}{5}\sum_{k=1}^n\frac{1}{5\cdot (n+1) + 1}.$$
A: By multiplying with $ 5(+5)$ we clear the denominators:
$- \frac{1(s+5)}{5s(s+5)}  + \frac{5}{5s (s+5)} = - \frac{1s}{5s(s + 5)}$ 
$-s = -s$
Which proves the induction step.
A: 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}
