Find a closed formula (not including $\sum$) for the expression $\sum_{k=0}^{n-1}\binom{2n}{2k+1}$

Question

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!

Answer 1

$$\sum_{k=0}^{2n}\binom{2n}k=(1+1)^{2n}=2^{2n}$$ $$\sum_{k=0}^{2n}(-1)^k\binom{2n}k=(1-1)^{2n}=[n=0]$$ Thus, subtracting the second sum from the first cancels all even indices and leaves only twice the odd indices: $$\sum_{k=0}^{n-1}\binom{2n}{2k+1}=\frac{2^{2n}-[n=0]}2=2^{2n-1}[n>0]$$

Answer 2

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}