prove that $2^{n-1}=\sum_{x=1}^{\frac{\left(n+\operatorname{mod}(n,2)\right)}{2}}\left(_n\mathrm{C}_r(n,2x-1)\right)$

Question

can you prove that $2^{n-1} = \sum_{x=1}^{\frac{\left(n+\operatorname{mod}(n,2)\right)}{2}}\left(\operatorname{nCr}\left(n,2x-1\right)\right)$.original how i got this was from the relation from (1+a)^n->(1-a)^n you can get this from $\left(1+a\right)^n-2\sum_{x=1}^{\frac{\left(n+\operatorname{mod}\left(n,2\right)\right)}{2}}\left(\operatorname{nCr}\left(n,2x-1\right)\cdot a^{\left(2x-1\right)}\right)=(1-a)^{n}$ and i just replaced a with 1 to get $2^n=2\sum_{x=1}^{\frac{\left(n+\operatorname{mod}\left(n,2\right)\right)}{2}}\left(\operatorname{nCr}\left(n,2x-1\right)\right)$ now just divide by 2 to get 2^(n-1)= $\sum_{x=1}^{\frac{\left(n+\operatorname{mod}\left(n,2\right)\right)}{2}}\left(\operatorname{nCr}\left(n,2x-1\right)\right)$ but i don't understand why this works.

Answer 1

Binomial expansion gives \begin{eqnarray*} (1+a)^n= \sum_{i=0}^{n} \binom{n}{i} a^i \\ (1-a)^n= \sum_{i=0}^{n} (-1)^i \binom{n}{i} a^i \\ \end{eqnarray*} Subtract these equations and observe that the "even" terms cancel, giving \begin{eqnarray*} (1+a)^n -(1-a)^n= 2\sum_{i=1}^{\lceil \frac{n}{2} \rceil } (-1)^i \binom{n}{2i-1} a^i \\ \end{eqnarray*} Now set $a=1$ and divide by $2$.