The Question:
$$ \mbox{To prove that:}\quad \frac{3!}{2(n+3)} = \sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}} $$
My Attempt:
I start off by writing $\sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}$ as $\sum_{r=0}^{n}{(-1)^r\frac{n!3!}{(n-r)!(r+3)!}}$.
Now, since there is a term of $3!$ in it, I thought it would be a good idea to convert the term inside the expression to a $\binom{n+3}{r+3}$ term (by multiplying and dividing by $(n+1)(n+2)(n+3)$ i.e.
$\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}{(-1)^r\frac{(n+3)!}{(n-r)!(r+3)!}}$.
This becomes,
$\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}{(-1)^r\binom{n+3}{r+3}}$
Beyond this, I absolutely have no clue. I have worked at this for hours but I still can't seem to get an alternative to this method so, I tried sticking to it but unfortunately, I couldn't come up with anything. I just can't seem to figure out what I can do to further simplify this expression!!Any hint on how to progress will be greatly appreciated.
Thanks!