$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction)

Question

I'm having some difficulty proving by induction the following statement.

$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$

I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) , but I am getting stuck on the induction step.

As far as I know I have to show $$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$ implies $$\sum_{i=0}^{n+1} \frac{1}{(i+3)(i+4)} = \frac{n+1}{4(n+5)}$$

To do this I think I should add the number $\frac{1}{(n+4)(n+5)}$ to $\frac{n}{4(n+4)}$ and see if it gives $\frac{n+1}{4(n+5)}$ , if I am not mistaken.

When trying to do that however I get stuck. I have:

$$\frac{n}{4(n+4)} +\frac{1}{(n+4)(n+5)} = \frac{n(n+4)(n+5)}{4(n+4)^2(n+5)} + \frac{4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+4)(n+5)+4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+5)+4}{4(n+4)(n+5)}$$

However beyond this point I don't know how to reach $\frac{n+1}{4(n+5)}$ I always just end up at the starting point of that calculation.

So I think that either my approach must be wrong or I am missing some trick how to simplify $$\frac{(n(n+5)+4}{4(n+4)(n+5)}$$

I would be very grateful for any help, as this is a task on a preparation sheet for the next exam and I don't know anyone, that has a correct solution.

Answer 1

$$\frac{n(n+5)+4}{4(n+4)(n+5)}=\frac{n^2+5n+4}{4(n+4)(n+5)}=\frac{(n+4)(n+1)}{4(n+4)(n+5)}=\frac{n+1}{4(n+5)}$$

Answer 2

$$\dfrac{n(n+5)+4}{4(n+4)(n+5)}=\dfrac{n^2+5n+4}{4(n+4)(n+5)}=\dfrac{(n+1)(n+4)}{4(n+4)(n+5)}$$

This simplifies to $\dfrac {n+1}{4(n+5)}$ and you want to substitute $n$ for $n+1$ for $n$.

Answer 3

$$\frac{n}{4(n+4)} +\frac{1}{(n+4)(n+5)} = \frac{n(n+4)(n+5)}{4(n+4)^2(n+5)} + \frac{4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+4)(n+5)+4(n+4)}{4(n+4)^2(n+5)} = \frac{(n(n+5)+4}{4(n+4)(n+5)}$$

Last expression can be simplified by $n+4$

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