Evaluate $\binom{12}0+\binom{12}2+\ldots+\binom{12}{12}$ using binomial theorem 
Solve the sum:
  $$ 
{12 \choose 0}+
{12 \choose 2}+
{12 \choose 4}+
{12 \choose 6}+
{12 \choose 8}+
{12 \choose 10}+
{12 \choose 12}
$$
using the binomial theorem.

I know the binomial theorem:
$$ \left(a+b\right)^2 = \sum_{k=0}^{n} {n \choose k} a^nb^{n-k}$$
However I fail to translate the sum into the theorem. I probably need to choose a variable ($b$?) to be negative to get half of the terms to disappear.
 A: With $a = 1, b = -1$ you get that
$$
\sum_{i= 0}^{12}(-1)^i\binom{12}{i} = (1-1)^{12} = 0
$$
and $a = 1, b = 1$ gives you
$$
\sum_{i = 0}^{12}\binom{12}{i} = (1 + 1)^{12} = 2^{12}
$$
Add these two together and see what terms are left.
Edit
Expanding, another way to write the $b = -1$ equation over is
$$
0 =  \binom{12}{0} -\binom{12}{1} +\binom{12}{2} -\binom{12}{3}+ \binom{12}{4}- \binom{12}{5}+ \binom{12}{6}\\- \binom{12}{7} +\binom{12}{8} -\binom{12}{9} +\binom{12}{10} -\binom{12}{11} +\binom{12}{12}
$$
The $b = 1$ equation ca be written
$$
2^{12}=  \binom{12}{0} +\binom{12}{1} +\binom{12}{2} +\binom{12}{3}+ \binom{12}{4}+ \binom{12}{5}+ \binom{12}{6}\\+ \binom{12}{7} +\binom{12}{8} +\binom{12}{9} +\binom{12}{10}+\binom{12}{11} +\binom{12}{12}
$$
If we add these two together, we get
$$
2^{12} = 2\binom{12}{0} +2\binom{12}{2} +2\binom{12}{4} +2\binom{12}{6}+ 2\binom{12}{8}+ 2\binom{12}{10}+ 2\binom{12}{12}
$$
Dividing by $2$ gets you the answer you're after.
A: Hint : let $\displaystyle S=\sum_{k=0}^n \binom{n}{2k}$ and 
$\displaystyle T=\sum_{k=0}^n \binom{n}{2k+1}$. Show that
$S+T=(1+1)^n$ and $S-T=(1-1)^n$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
\begin{align}
&{\large\sum_{k = 0}^{6}{12 \choose 2k}}
=
\sum_{k = 0}^{5}{11 \choose 2k} + \sum_{k = 1}^{6}{11 \choose 2k - 1}
=
\sum_{k = 0}^{11}{11 \choose k} = {\large 2^{11}}
\end{align}
