Let $M$ be a positive integer. Then I believe that the following fun identity is true:
$$\sum_{k=0}^{\text{Floor}(M/2)}2^{M-2k}{M\choose 2k}{2k\choose k} = {2M\choose M}$$
Numerically it checks out for $M$ up to 100, but of course I would like a real proof. Unfortunately none of the standard manipulations I'm familiar with for binomial coefficients have gotten me anywhere - I thought replacing the exponential with a sum over binomial coefficients might help, but so far, nothing. Induction seems difficult too. Any suggestions?
Generating functions to the rescue! It suffices to multiply both sizes by $x^M$, sum over nonnegative integers $M$, and confirm that the two answers are the same. We use the known identities $$ \sum_{k=0}^\infty \binom{2k}k z^k = \frac1{\sqrt{1-4z}} \quad\text{and}\quad \sum_{M=j}^\infty \binom Mj y^M = \frac{y^j}{(1-y)^{j+1}}. $$ The first identity suffices to handle the right-hand side: $$ \sum_{M=0}^\infty \binom{2M}M x^M = \frac1{\sqrt{1-4x}}. $$ For the left-hand side: \begin{align*} \sum_{M=0}^\infty x^M \sum_{k=0}^{\lfloor M/2\rfloor}2^{M-2k}{M\choose 2k}{2k\choose k} &= \sum_{k=0}^\infty 4^{-k} {2k\choose k} \sum_{M=2k}^\infty (2x)^M {M\choose 2k} \\ &= \sum_{k=0}^\infty 4^{-k} {2k\choose k} \frac{(2x)^{2k}}{(1-2x)^{2k+1}} \\ &= \frac1{1-2x} \sum_{k=0}^\infty {2k\choose k} \biggl( \frac x{1-2x} \biggr)^{2k} \\ &= \frac1{1-2x} \frac1{\sqrt{1-4\{x/(1-2x)\}^2}} = \frac1{\sqrt{1-4x}} \end{align*} as desired.