Let the left side of the equation be $a_n$. Then, we have $$\begin{align*} a_n =\ & 2^{2n} + \sum_{k=1}^{n}(-1)^k2^{2n-2k}\left(\binom{2n-k}{k}+\binom{2n-k}{k-1}\right) \\ =\ &\sum_{k=0}^{n}(-1)^k2^{2n-2k}\binom{2n-k}{k}+ \sum_{k=1}^{n}(-1)^k2^{2n-2k}\binom{2n-k}{k-1} \end{align*}$$

Let the $1^{st}$ term be $b_n$. And, replacing $(k-1)$ with $l$ in the $2^{nd}$ term leads to $$\sum_{k=1}^{n}(-1)^k2^{2n-2k}\binom{2n-k}{k-1}=\sum_{l=0}^{n-1}(-1)^{l+1}2^{2(n-1)-2l}\binom{2(n-1)-l+1}{l} =-a_{n-1}$$

$$\therefore\quad a_n = b_n - a_{n-1} \implies b_n = a_n + a_{n-1}$$

Meanwhile, we have $$\begin{align*} b_n =\ &\sum_{k=0}^{n}(-1)^k2^{2n-2k}\binom{2n-k}{k} \\ =& 2^{2n} + \sum_{k=1}^{n-1}(-1)^k2^{2n-2k}\binom{2n-k}{k} + (-1)^n \\ =\ & 2^{2n} + \sum_{k=1}^{n-1}(-1)^k2^{2n-2k}\left(\binom{2n-k-1}{k}+\binom{2n-k-1}{k-1}\right) + (-1)^n \\ =\ & \sum_{k=0}^{n-1}(-1)^k2^{2n-2k}\binom{2n-k-1}{k} + \sum_{k=1}^{n}(-1)^k2^{2n-2k}\binom{2n-k-1}{k-1}  \\ =\ & 4 \sum_{k=0}^{n-1}(-1)^k2^{2(n-1)-2k}\binom{2(n-1)-k+1}{k} \\ &+ \sum_{l=0}^{n-1}(-1)^{(l+1)}2^{2(n-1)-2l}\binom{2(n-1)-l}{l}\\=\ & 4a_{n-1} - b_{n-1} \end{align*}$$

Hence, we have $$\left\{ \begin{array}{rl} b_n = & a_n + a_{n-1} \\ b_n=& 4a_{n-1} - b_{n-1} \end{array} \right. \implies a_n=2a_{n-1}-a_{n-2}$$

This is a standard linear sequence which can be rewritten as $$a_n - a_{n-1} = a_{n-1}-a_{n-2}$$

Because $a_0 = 1$ and $a_1 =2$, we have $$\begin{array}{rl} & a_n - a_{n-1} = a_{n-1} - a_{n-2} = \cdots = a_1 - a_0 = 1 \\ \implies & a_n = a_{n-1} + 1 = a_{n-2}+2 = \cdots = a_0 + n = n+1 \end{array}$$