Prove $\binom{n}{0}$ + $\binom{n}{2}$ + $\binom{n}{4} + ...+\binom{n}{n-1}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+...+\binom{n}{n}$ I want to prove that $\binom{n}{0}$ + $\binom{n}{2}$ + $\binom{n}{4} + ...+\binom{n}{n-1}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+...+\binom{n}{n}$ for all odd numbers $n \ge 1$
I see that $\binom{n}{0}$ and $\binom{n}{n}$ are both equal to $1$ and therefor remove each other.
But what can I do after that? I'm very lost, please help me move in the right direction at least.

You can prove by induction on $n$ that
$$\sum_{k} (-1)^k \binom{n}{k} =0$$
for all $n\ge 1$. For $n=1$ it's easy to check. Assume that it's true for some $n\ge 1$. Denote by $a_k = \binom{n}{k}$, $a_k'= \binom{n+1}{k}$. So we have
$$\sum_{k}(-1)^k a_k = 0$$
and want to show that
$$\sum_{k} (-1)^k a_k'=0$$
We have (Pascal triangle)
$$a_{k}' = a_k+ a_{k-1}$$
Now break the sum $\sum_k (-1)^k a_k'$ into two sums.
$\bf{Added:}$
In a similar way one proves by induction on $n$ that
$$\sum_k \binom{n}{k} = 2^n$$
or even
$$\sum_k x^k \binom{n}{k} = (1+x)^n$$
