proof of combinatoric/using pascals theorem prove that, for even values of $n$, $$\sum_{i=0}^{n/2}\binom{n}{2i}= 2^{n-1}\;.$$
I tried using pascals theorem to help prove this with no success
 A: You want to prove that
$$\sum_{i=0}^{n/2}\binom{n}{2i}=2^{n-1}\tag{1}$$
when $n$ is even. The easiest way is with a combinatorial argument. The lefthand side of $(1)$ counts the subsets of $\{1,\dots,n\}$ of even cardinality. Exactly half of the subsets of any finite, non-empty set have even cardinality, and $\{1,\dots,n\}$ has $2^n$ subsets altogether, so ... ?
Added: You can use Pascal’s identity to prove it, though.
$$\begin{align*}
\sum_{i=0}^{n/2}\binom{n}{2i}&=\sum_{i=0}^{n/2}\left(\binom{n-1}{2i-1}+\binom{n-1}{2i}\right)\tag{2}\\\\
&=\sum_{i=0}^{n-1}\binom{n-1}i\\\\
&=2^{n-1}\;.
\end{align*}$$
The terms on the righthand side of $(2)$ actually range from $\binom{n-1}{-1}$ through $\binom{n-1}n$, but the two extra terms, $\binom{n-1}{-1}$ and $\binom{n-1}n$, are both $0$, so I simply dropped them in the single summation in the next line.
A: Let
$$
E:=\sum_{0\le i\le n,\ \  i \text{  even}} \binom{n}{i},\ \ \ \ \ 
O:=\sum_{0\le i\le n,\ \  i \text{  odd}} \binom{n}{i}.
$$
By the binomial theorem, 
$$
E+O=\sum_{0\le i\le n} \binom{n}{i}=(1+1)^n=2^n$$
and, if $n\ge 1$,
$$
E-O=\sum_{0\le i\le n} (-1)^i \binom{n}{i}=(1-1)^n=0.$$
Therefore, $E=O=2^{n-1}$.
A: Alternating Sum
If we take the alternating sum of any row other than the top row we get something like the following:
$\hspace{2cm}$
Each number in gray contributes to one number in the lower row which is positive in the sum (green +) and one that is negative (red -) in the sum. Thus, the alternating sum adds and subtracts each of the gray numbers, giving a total of $0$.
Non-Alternating Sum
Each number in gray contributes to two numbers in the lower row. Thus the sum of the lower row is twice that of the upper row.
Putting It Together
From the arguments above, we see that for all but the top row,
$$
\sum_i\binom{n}{2i}-\sum_i\binom{n}{2i+1}=0\tag{1}
$$
and since the sum of the $n=0$ row is $1$, we get
$$
\sum_i\binom{n}{2i}+\sum_i\binom{n}{2i+1}=2^n\tag{2}
$$
Adding $(1)$ and $(2)$ and dividing by $2$ yields
$$
\sum_i\binom{n}{2i}=2^{n-1}\tag{3}
$$
