TITLE: A combinatorial identity with binomial coefficients and floor function.
QUESTION [9 upvotes]: Show that $$ \sum_{m=0}^{2k+1}{2^{m}\binom{n}{m}\binom{n-m}{\bigl\lfloor \frac{2k+1-m}{2} \bigr\rfloor}}=\binom{2n+1}{2k+1} $$
I tried expending the sum and using induction, but could not complete the induction step; I tried using Pascal's identity to obtain $$ \binom{n-m}{\bigl\lfloor \frac{2k+3-m}{2} \bigr\rfloor}=\binom{n-m+1}{\bigl\lfloor \frac{2k+3-m}{2} \bigr\rfloor}-\binom{n-m}{\bigl\lfloor \frac{2k+1-m}{2} \bigr\rfloor} $$ but couldn't find other identities to complete the induction step. Searching Gould's Combinatorial Identities brought up nothing, though I could easily had missed something useful there.
I'm looking to complete the induction step, but would also like to find an algebraic or a combinatorial proof. I would like to avoid using trigonometric and root functions, but would like to use complex numbers in cartesian form.

REPLY [8 votes]: Here is a combinatorial proof. The right hand side counts the number of ways to choose $2k+1$ elements from a set with $2n+1$ elements. Fix a partition of the latter set into $n+1$ disjoint subsets, namely a singleton and $n$ two-element subsets.
In order to choose such $2k+1$ elements, first we choose the number $m$ of two-element subsets from which we will select exactly one element. Now we choose such subsets, which can be made in $\binom nm$ ways. Finally, from each one of these $m$ subsets we choose exactly one element; this can be made in $2^m$ ways.
It remains to choose the remaining $2k+1-m$ elements among the elements of the singleton and the remaining $n-m$ two-element members of the partition. Because of the previous step, picking an element of one of these two-element subsets forces us to pick the other element as well. Thus, by choosing $\ell$ of these $n-m$ two-element subsets (which can be made in $\binom{n-m}\ell$ ways) we are adding $2\ell$ elements to our $m$ already chosen elements. In other words, at this point we already chose $2\ell+m$ elements.
The important thing to note is that at this point the value of $\ell$ is determined, that is, we cannot "choose" such value. In fact: if $m$ is even, then we necessarily must pick up the element from the singleton subset, and equality $2\ell+m+1=2k+1$ implies $\ell=\frac{2k-m}2=\bigl\lfloor\frac{2k+1-m}2\bigr\rfloor$; if $m$ is odd then we cannot choose the element from the singleton, and in this case $2\ell+m=2k+1$ implies $\ell=\frac{2k+1-m}2=\bigl\lfloor\frac{2k+1-m}2\bigr\rfloor$.