<s>[INST] Verifying Touchard's Identity $$C_{n+1} = \sum_{k=0}^{\lfloor n/2\rfloor}{n\choose 2k}\cdot C_k\cdot 2^{n-2k}$$
where $C_n$ are the Catalan numbers.
I think we start by diving both sides by $2^n$, but unsure of where to go from there
 [/INST] Notice that $2^{-2k}\binom{n}{2k}\binom{2k}{k}=\binom{n/2}{k}\cdot \binom{n/2-1/2}{k}$. So
$$A_n=\sum_{k=0}^{\lfloor n/2 \rfloor}2^n \cdot 2^{-2k}\cdot \binom{n}{2k}\binom{2k}{k}\cdot\dfrac{1}{k+1}=2^n\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n/2}{k}\cdot \binom{n/2-1/2}{k}\cdot\dfrac{1}{k+1}$$
Without loss of generality, suppose $n/2 \in \mathbb{N}$. By Vandermonde's identity, we have
$$A_n=\dfrac{2^n}{n/2+1} \cdot \binom{n+1/2}{n/2}=\dfrac{1}{n+2}\binom{2(n+1)}{n+1}=C_{n+1}$$
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