0
$\begingroup$

Anyone has idea how to simplify this summation: $$\sum\limits_{k=0}^{N}\binom{M-1+k}{k}\binom{M-1+N-k}{N-k}$$

$\endgroup$

marked as duplicate by Dietrich Burde, user186170, user91500, Aaron Maroja, Pragabhava Oct 13 '16 at 15:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

Note that for $|x|<1$, and for any positive integer $m$, $$\frac{1}{(1-x)^{m}}=\sum_{i=0}^{\infty}\binom{-m}{i}(-x)^i =\sum_{k=0}^{\infty}\binom{m-1+k}{k}x^k.$$ Hence, by the Cauchy product of series, $$\sum\limits_{k=0}^{N}\binom{M-1+k}{k}\binom{M-1+N-k}{N-k}$$ is the coefficient of $x^N$ of the product $$\frac{1}{(1-x)^{M}}\cdot \frac{1}{(1-x)^{M}}=\frac{1}{(1-x)^{2M}}.$$ By the above formula, such coefficient is equal to $$\binom{2M-1+N}{N}.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.