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!
$$\frac{k+1}{k+1-i} {k+1-i \choose i} = \frac{k}{k-i} {k-i \choose i} + \frac{k-1}{k-i} {k-i-1 \choose i-1}$$
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?
