Here's a part of the Pascal triangle:

Rows of this triangle are numbered from $0$, and the sum of $n$-th row is $\color{red}{2^n}$, because numbers in such row gives altogether count of all subsets of an $n$-element set, i. e. the number of elements in the power set of an $n$-element set.
Now note odd-numbered rows, e. g. the last one $\, (\text{the }5^\text{th})$. They have even number of elements, and the first is the same as the last, the second is the same as the last but one, etc.
because of the know relation
$$\quad{n \choose k} = {n \choose {n-k}}$$
giving us
\begin{aligned}
{5 \choose 0} &= {5 \choose 5} \\[1ex]
{5 \choose 1} &= {5 \choose 4} \\[1ex]
{5 \choose 2} &= {5 \choose 3} \\
\end{aligned}
Because in these equivalent pairs there is $1$-$1$ mapping between the even and odd "bottom" numbers
$$0 \mapsto 5\\ 2 \mapsto 3\\ 4 \mapsto 1$$
the sum for the even, and and the sum for the odd ones must be the same, so it is a half of the total sum (for even and odd members):
$$\color{black}{{2^5\over 2} = 2^4 = 16}$$
for our particular number $5 = 2n+1$. For general solution we will use $(2n+1)$ instead of $5$, obtaining the result
$$\color{red}{{2^{2n+1}\over 2} = 2^{2n}}$$