<s>[INST] Proving that $\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+\cdots+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$ by induction I've proved the base case where $n=1$ and made the assumption that $n=k$ is true, but I'm stuck on the $n=k+1$ part.  I just cannot seem to get the algebra to work in my favor.  
Here is the original:

$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+…+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}\qquad \forall\, n \in \mathbb N$

I get it to the following form and just run in circles:

$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+...+\frac{1}{(k+3)(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$

Simplifying

$\displaystyle\frac{k}{4(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$

Is this in the correct form?
 [/INST] This should help:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{(i+3)(i+4)}&= \sum_{i=1}^k\frac{1}{(i+3)(i+4)}+\frac{1}{(k+4)(k+5)}\\[1em]
&= \frac{k}{4(k+4)}+\frac{1}{(k+4)(k+5)}\\[1em]
&= \frac{k(k+5)+4}{4(k+4)(k+5)}\\[1em]
&= \frac{(k+1)(k+4)}{4(k+4)(k+5)}\\[1em]
&= \frac{k+1}{4(k+5)}.
\end{align}
See if you can spot where the inductive hypothesis is used and where some basic algebraic manipulation plays an important role.
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