<s>[INST] Find a closed formula (not including $\sum$) for the expression $\sum_{k=0}^{n-1}\binom{2n}{2k+1}$ Find a closed formula (not including $\sum$) for the expression
$$\sum_{k=0}^{n-1}\binom{2n}{2k+1}$$
I started by using the fact that
$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$
to get that
$$\sum_{k=0}^{n-1}\binom{2n}{2k+1}=\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}+\binom{2n-1}{2k}$$
$$=\sum_{k=0}^{n-1}\binom{2n-2}{2k+1}+\binom{2n-2}{2k}+\binom{2n-2}{2k}+\binom{2n-2}{2k-1}$$
now letting $m=n-1$
$$\sum_{k=0}^{m}\binom{2m+2}{2k+1}=\sum_{k=0}^{m}\binom{2m}{2k+1}+2\cdot\binom{2m}{2k}+\binom{2m}{2k-1}$$
I'm not exactly sure where to go from here or even if this has been helpful. Any guidance/alternative methods would be greatly appreciated!
 [/INST] Note that
$$\sum_{k\ge0} a_{2k+1} = \sum_{k\ge0} \frac{1-(-1)^k}{2}a_k.$$
Taking $$a_k=\binom{2n}{k}$$ yields
\begin{align}
\sum_{k\ge0} \binom{2n}{2k+1} 
&= \sum_{k\ge0} \frac{1-(-1)^k}{2} \binom{2n}{k} \\
&= \frac{1}{2}\sum_{k\ge0} \binom{2n}{k} - \frac{1}{2}\sum_{k\ge0} (-1)^k\binom{2n}{k} \\
&= \frac{1}{2}(1+1)^{2n} - \frac{1}{2}(1-1)^{2n} \\
&= \frac{4^n-[n=0]}{2}.
\end{align}
</s>