Proving the combinatorial identity $\sum_{k=0}^{n} \binom{2n+1}{2k}=2^{2n}.$ I want to see that 
$$\sum_{k=0}^{n} \binom{2n+1}{2k}=2^{2n}.$$
Does anybody know how to see this quickly?
 A: By the binomial theorem, one has
$$2^m=(1+1)^m=\sum_{k=0}^{m}\binom{m}{k}\tag 1$$
and
$$0=(1-1)^m=\sum_{k=0}^{m}\binom{m}{k}(-1)^k\tag 2$$
Now $(1)+(2)$ gives you
$$2^m=2\left(\binom{m}{0}+\binom{m}{2}+\binom{m}{4}+\cdots\right).$$
A: Hint: Use a) $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$
b) $\sum_{k=0}^n \binom{n}{k} = 2^n$
A: You can use a pascal triangle!
I shall use n = 2 as a example first.
For this, we take the 5th line, which is 1 5 10 10 5 1.
2k chooses the numbers in the odd even positions, which are in this case:
1, 10, 5
This can work for all other lines as well.
The sum of this is half of the sum of the whole line, which is $2^{2n+1}$
$\frac{1}{2} \times 2^{2n+1} = 2^{2n}$
A: Let $S=\{1, \cdots ,2n\}$ and $T=\{1, \cdots, 2n, 2n+1\}$.
We can set up a bijection between the subsets of S and the even-numbered subsets of T  
by mapping $A\rightarrow\begin{cases} A &\mbox{,   if |A| is even}\\A\cup\{2n+1\}&\mbox{, if |A| is odd}\end{cases}$.
Therefore $\displaystyle 2^{2n}=\sum_{k=0}^{n}\binom{2n+1}{2k}$,
since S has $2^{2n}$ subsets and T has $\displaystyle\sum_{k=0}^{n}\binom{2n+1}{2k}$ subsets with an even number of elements.
