Evaluation of a tricky binomial sum

Question

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!

Answer 1

Start from $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}(-1)^r \binom{n+3}{r+3} $$Reindex: $$ =\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r-3} \binom{n+3}{r} $$ $$ =\frac{-3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r} \binom{n+3}{r} $$Now add and subtract the terms $0\leq r\leq 2$: $$ =\frac{-3!}{(n+1)(n+2)(n+3)}\left(\sum_{r=0}^{n+3}(-1)^{r} \binom{n+3}{r}-\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right) $$The binomial theorem graciously takes care of the first series for us, and then the result falls out: $$ =\frac{3!}{(n+1)(n+2)(n+3)}\left(0+\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right) $$ $$ =\frac{3!}{(n+1)(n+2)(n+3)}\cdot \frac{(n+1)(n+2)}{2} = \frac{3}{n+3} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{r = 0}^{n}\pars{-1}^{r}\,{{n \choose r} \over {r + 3 \choose r}} & = 3\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r} {\Gamma\pars{r + 1}\Gamma\pars{3} \over \Gamma\pars{r + 4}} \\[5mm] & = 3\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r} \int_{0}^{1}t^{r}\pars{1 - t}^{2}\,\dd t \\[5mm] & = 3\int_{0}^{1}\bracks{\sum_{r = 0}^{n}{n \choose r} \pars{-t}^{r}}\pars{1 - t}^{2}\,\dd t = 3\int_{0}^{1}\pars{1 - t}^{n}\pars{1 - t}^{2}\,\dd t \\[5mm] & =\left. -3\,{\pars{1 - t}^{n + 3} \over n + 3}\right\vert_{0}^{1} = {3 \over n + 3} = \bbx{3! \over 2\pars{n + 3}} \end{align}