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

Question

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.

Answer 1

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. $\frac{1}{2}\cdot 2^{2n+1} = 2^{2n}$.

Answer 2

\begin{align*}\sum_{k=0}^n\binom{2n+1}k&=\frac12\sum_{k=0}^n\left(\binom{2n+1}k+\binom{2n+1}{2n+1-k}\right)\\&=\frac12\sum_{k=0}^{2n+1}\binom{2n+1}k\\ &=\frac{2^{2n+1}}2=2^{2n}.\end{align*}

Answer 3

Hint: Use binomial expansion for $(1+1)^{2n}$.