<s>[INST] 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 $\sum_\limits{k=0}^{2n}{2n \choose k}=\sum_\limits{k=0}^{n} {2n+1 \choose k}$?
 [/INST] 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: 
$$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k} = \frac{1}{2} \cdot 2^{2n + 1} = 2^{2n} = 4^n$$
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