# Combinatorics Proof for $\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$

I've tried to prove that $\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$ using combinatorics reasoning.

whereas $2^{2n}$ can be, for example, the number of options to give 2n students at max one cookie per student, I can't find an explantion for the right expression.

The LHS accounts for the subsets of $E=\{1,2,\ldots,2n+1\}$ whose cardinality is at most $n$.
For any subset $B\subseteq E$, either $B$ or (exclusive) $E\setminus B$ meets such constraint, hence the LHS equals half the number of subsets of $E$, i.e. $\frac{1}{2}\cdot 2^{2n+1} = 2^{2n}$.
Hint: Use binomial expansion for $(1+1)^{2n}$.