Evaluate ${k \choose 0} + {k \choose 2} + {k \choose 4}+ {k \choose 6} +...+ {k \choose k}$ for even $k$

Question

I want to evaluate ${k \choose 0} + {k \choose 2} + {k \choose 4}+ {k \choose 6} + ... +{k \choose k}$ for even $k$ in closed form, using methods of counting. How could this be done?

Answer 1

Hint for a combinatorial proof.

Obviously this is counting number of even-sized subsets of a set of $k$ objects.

When $k$ is odd, can you see what ${k \choose 0} + {k \choose 2} + \cdots$ should be, without explicit calculation?

Then when $k$ is even, consider one specific object in the set. That object can be either in the subset or not. In both cases the problem reduces to choosing the rest of the subset from the remaining $k-1$ objects.

Answer 2

For convenience, write $k=2t$. Then you want to evaluate $\sum_{i=0}^t \binom {2t}{2i}$. Now by the binomial theorem, $$(1+1)^{2t}=\sum_{i=0}^{2t}\binom{2t}i,$$ and $$(1-1)^{2t}=\sum_{i=0}^{2t}(-1)^i\binom{2t}i.$$

Adding these together gets rid of the odd terms, but then the even terms are doubled. Therefore the result is $$\frac{(1+1)^{2t}+(1-1)^{2t}}2=\boxed{2^{k-1}}.$$