Closed form of $\sum\limits_{k=0}^{n} {2n+1 \choose k}$ I am stuck with this problem:  $\sum_\limits{k=0}^{n} {2n+1 \choose k}$.
I used Wolframalpha for the answer and is $4^n$. So i searched for a similar way to expressed it which is:
$$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}{2n \choose k}$$
Is there a way to prove that $\sum_\limits{k=0}^{2n}{2n \choose k}=\sum_\limits{k=0}^{n} {2n+1 \choose k}$?
 A: Since $$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\left(\sum_{k = 0}^n \binom{2n + 1}{k} + \sum_{k = 0}^n \binom{2n + 1}{2n+1-k}\right) = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k}$$
Then we have: 
$$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k} = \frac{1}{2} \cdot 2^{2n + 1} = 2^{2n} = 4^n$$
A: I post this as an answer rather than a comment...
Note  that $ \binom{2n + 1}{k} =\binom{2n + 1}{2n+1-k} $
$$\sum_{k = 0}^n \binom{2n + 1}{k} =\sum_{k = 0}^n \binom{2n + 1}{2n+1-k} $$
$$\sum_{k = 0}^n \binom{2n + 1}{2n+1-k} =\sum_{k = n+1}^{2n+1} \binom{2n + 1}{k}  $$
Therefore
$$\sum_{k = 0}^n \binom{2n + 1}{k}=\sum_{k = n+1}^{2n+1} \binom{2n + 1}{k} $$
You can conclude that 
$$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac 1 2\sum_{k = 0}^n \binom{2n + 1}{k}+ \frac 1 2 \sum_{k = n+1}^{2n+1} \binom{2n + 1}{k} =\frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k}=4^n $$
A: There is a very nice way to think about this problem:
The summation represents the number of subsets of all sizes $i: i\leq n$ from a set $A$ of size $2n+1$. Now, $$\begin{align*} | \mathcal{P}(A)| &= 2^{2n+1} \\ &=2\cdot 4^n \end{align*}$$
Is the number of every subset of $A$
But notice exactly half of the subsets of $A$ are of sizes less than or equal to $n$, so we have $4^n$ subsets, as required.
