<s>[INST] 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.
 [/INST] 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
$$ \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)}=\frac{1}{5}\sum_{k=1}^n\left(\frac{1}{5k + 1}-\frac{1}{5(k+1) + 1}\right)=\frac{1}{5}\sum_{k=1}^n\frac{1}{5k + 1}-\frac{1}{5}\sum_{k'=2}^{n+1}\frac{1}{5k' + 1}\\=\frac{1}{5}\sum_{k=1}^n\frac{1}{5\cdot 1 + 1}
-\frac{1}{5}\sum_{k=1}^n\frac{1}{5\cdot (n+1) + 1}.$$
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