Proof $\displaystyle \sum_{k=0}^n \binom{2n}{2k} = 2^{2n-1}$

Question

Can someone help me prove the equation: $$\sum_{k=0}^n \binom{2n}{2k} = 2^{2n-1}$$

I know that via binomial theorem for even k: $$2^{2n} = (1+1)^{2n} = \sum_{k=0}^n \binom{2n}{k}$$ and for odd k: $$0 = (1-1)^{2n} = \sum_{k=0}^n \binom{2n}{k}\cdot(-1)^k$$ That's how i solve the equation: $$2^{2n-1}=\frac{2^{2n}}{2}=\frac{\sum_{k=0}^n \binom{2n}{k}}{2}$$ Now we can remove all odd k because they are equal 0. Also only even k remain: $$\sum_{k=0}^n \binom{2n}{2k}=2^{n+1}$$ And.. I have a mistake somewhere, because it must be: $$2^{n-1}$$ I think I don't fully understand the difference between odd k and the number of subsets with odd cardinality. Or odd k is an abbreviation for the number of subsets with odd cardinality?

Answer 1

You have $2n-1$ beans on the table and you want to pick some subset of them. Borrow another bean and color it red. Now pick a subset of the remaining $2n$ beans of even cardinality, and throw out the red one if it's in there.

Answer 2

I think you had a confusion when you took the case of k odd and k even, in both sums there are k even and odd I did not understand your remark however here is my attempt:

\begin{align*} 0 &= (1-1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k \\ \text{So:} \\ 0 &= \sum_{k=0}^{n} \binom{2n}{2k} - \sum_{k=0}^{n-1} \binom{2n}{2k+1} \\ \text{And:} \\ 2^{2n} &= (1+1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} \\ \text{Then:} \\ 2^{2n} &= \sum_{k=0}^{n} \binom{2n}{2k} + \sum_{k=0}^{n-1} \binom{2n}{2k+1} \\ \text{Therefore:} \\ 2^{2n} &= 2 \sum_{k=0}^{n} \binom{2n}{2k} \\ \text{Finally:} \\ 2^{2n-1} &= \sum_{k=0}^{n} \binom{2n}{2k} \end{align*}