# Binomial Coefficient Identity, Double Series, Floor Function

Show that for any natural numbers $m$ and $n$ such that $m \le n$ that: $$\sum_{i=0}^{n}{ \sum_{j=0}^{m}{ \left(-1 \right)^{\lfloor \frac{i}{2} \rfloor+j}2^{n-i}\binom{n-\lfloor \frac{i+1}{2} \rfloor}{j}\binom{n-\lfloor \frac{i+1}{2} \rfloor-j}{\lfloor \frac{i}{2} \rfloor}\binom{i}{m-j}}}= \left(-1 \right)^{m} \binom{2n+1}{2m+1}$$ I would like to avoid using generating functions but would like to use other techniques which regard coefficients of polynomials; I would like to avoid using calculus, trigonometric functions, chabyshev polynomials, complex numbers and mathematical induction. I would like to find an algebraic and combinatorial proofs.

After some work with binomial identities, the LHS becomes: $$\sum_{i=0}^{n}{ \left(-1 \right)^{\lfloor \frac{i}{2} \rfloor}2^{n-i}\binom{n-i+\lfloor \frac{i}{2} \rfloor}{n-i} \left(\sum_{j=0}^{m}{ \left(-1 \right)^{j}\binom{n-i}{j}\binom{i}{m-j}}\right)}$$ This form seems like it should be quite usefull but I haven't found the right binomial identities to keep going.

Update: cross posted to MO