How to prove that subset at odd size is equal to subset at even size?

Question

I saw this question: The number of odd size subsets is equal to the number of even size subsets but I'm trying to prove it at a different way:
I'm trying to prove that: $$\sum_{i=0}^{\left\lceil \frac{n}{2}\right\rceil }{n \choose 2i}=\sum_{i=0}^{\left\lceil \frac{n}{2}\right\rceil }{n \choose 2i+1}$$

if $n$ is odd, I can do it with the identity: ${n \choose k}={n \choose n-k}$, because if we have odd number like $7$ we have $4$ ($=\left\lceil \frac{n}{2}\right\rceil$) pairs: $(7,0),(6,1),(5,2),(4,3)$. The left element is odd and the right one is even. So:
$${7 \choose 0}={7 \choose 7},\,{7 \choose 6}={7 \choose 1},...$$

But when I'm trying to do this about even numbers (like $8$), I can't use this method.
So my question is: What can I do at cases of even numbers?

Thank you!

Answer 1

The claim is only true for non-empty sets $A$. There's a reason why the observation that (for any fixed element $a\in A$) $$ X\mapsto X\operatorname\Delta\{a\}$$ is a bijection from the even-sized subsets of $A$ to the odd-sized subsets (as well as vice versa) is considered a simpler proof of the desired claim.

Answer 2

I shall show how to prove this in the case $n=8$, and it should be clear how this idea generalizes. You need to prove $$ \binom80+\binom82+\binom84+\binom86+\binom88\stackrel{?}=\binom81+\binom83+\binom85+\binom87 $$ First, apply Pascals rule to each entry in the above purported equation except for $\binom80$ and $\binom88$, which says that $\binom{8}{k}=\binom{7}k+\binom7{k-1}$. The result is $$ \begin{align} \binom80+\left[\binom71+\binom72\right]+\left[\binom73+\binom74\right]+\left[\binom75+\binom76\right]+\binom88=\\ \left[\binom70+\binom71\right]+\left[\binom72+\binom73\right]+\left[\binom74+\binom75\right]+\left[\binom76+\binom77\right] \end{align} $$ After realizing that $\binom80=\binom70$ and $\binom88=\binom77$, the last equation should be obvious.