I have been working on proving the following combinatorial identity:
$$\frac{k+1}{k+1-i} {k+1-i \choose i} = \frac{k}{k-i} {k-i \choose i} + \frac{k-1}{k-i} {k-i-1 \choose i-1}$$
and I would appreciate any guidance or insights.
I tried applying some well-known combinatorial identities such as Pascal's Identity and Vandermonde's Identity, but I couldn't figure out how to manipulate them to obtain the desired result.
Or maybe this equation is not true?
Thank you!
The identity does not hold true in general, as noted in a comment already.
This looks related to OP's other question, which reduces to the slightly different (correct) identity:
$$\frac{k+1}{k-i+1} {k-i+1 \choose i} = \frac{k}{k-i} {k-i \choose i} + \frac{k-1}{k-i} {\color{red}{k-i} \choose i-1}$$
This combinatorial identity can be proved directly using that $\,\binom{m}{n} = \frac{m(m-1)\ldots(m-n+1)}{n!}\,$:
$$ \require{cancel} \frac{k+1}{\xcancel{k-i+1}} \cdot\frac{\xcancel{(k-i+1)} \cdot \cancel{(k-i)\ldots(k-2i+2)}}{i \cdot \bcancel{(i-1)!}} \\= \frac{k}{k-i} \cdot\frac{\cancel{(k-i)\ldots(k-2i+2)} \cdot (k-2i+1)}{i \cdot \bcancel{(i-1)!}} + \frac{k-1}{k-i} \cdot \frac{\cancel{(k-i)\ldots(k-2i+2)}}{\bcancel{(i-1)!}} \\ \iff\qquad \frac{k+1}{i} = \frac{k(k-2i+1)}{(k-i)i} + \frac{k-1}{k-i} \\ \iff\qquad (k+1)(k-i) = k(k-2i+1) +(k-1)i \qquad\iff\qquad \ldots $$
Other proofs can be found under the Has anyone studied this binomial like coefficients before?