Let $a_n=\displaystyle\sum_{k=0}^{n}\binom{2n}{k}$ and $b_n=\displaystyle\sum_{k=n}^{2n}\binom{2n}{k}$. Because $\displaystyle\binom{2n}{k}=\displaystyle\binom{2n}{2n-k}$ holds for every $0\le k\le 2n$, it must be true that $a_n=b_n$. Meanwhile, $$a_n+b_n=\sum_{k=0}^{2n}\binom{2n}{k}+\binom{2n}{n}=2^{2n}+\binom{2n}{n}$$

Setting $a_n=b_n$ and dividing both sides by $2$ immediately lead to the desired result.