Evaluate $\binom{12}0+\binom{12}2+\ldots+\binom{12}{12}$ using binomial theorem

Question

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.

Answer 1

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.

Answer 2

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 $S-T=(1-1)^n$.

Answer 3

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\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}