$$ \mbox{If}\quad s_{n} = \sum_{k = 0}^{n}\left(-4\right)^{k} \binom{n + k}{2k},\quad\mbox{how to prove}\quad s_{n + 1} + 2s_{n} + s_{n - 1} = 0\ ?. $$
It would be great if someone helps.
Let us prove a stronger result: for $n\ge1$ $$ s_n=\sum_{k=0}^{n}(-4)^k\binom{n+k}{2k}=(-1)^n(2n+1),\quad\tag1 c_n=\sum_{k=0}^{n}(-4)^k\binom{n+k-1}{2k-1}=(-1)^n 4n. $$
Indeed for $n=1$ the formula $(1)$ is valid as can be easily checked. Assume it is valid for some $n\ge1$. Then it is valid for $n+1$ as well: $$ \begin{align} c_{n+1}&=\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1}\\ &=\sum_{k=0}^{n+1}(-4)^k\left[\binom{n+k-1}{2k-1}+\binom{n+k-1}{2k-2}\right]\\ &=c_n-4s_n\\& =(-1)^n4n-(-1)^{n}4(2n+1)\\ &=(-1)^{n+1}4(n+1);\\ \vphantom{X}\\ s_{n+1}&=\sum_{k=0}^{n+1}(-4)^k\binom{n+1+k}{2k}\\ &=\sum_{k=0}^{n+1}(-4)^k\left[\binom{n+k}{2k}+\binom{n+k}{2k-1}\right]\\ &=s_n+c_{n+1}\\& =(-1)^n(2n+1)+(-1)^{n+1}4(n+1)\\ &=(-1)^{n+1}(2(n+1)+1). \end{align}$$
Thus by induction the formula $(1)$ is proved. The original statement is a trivial corollary.
This answer proves that $s_{n+1}+2s_n+s_{n-1}=0$ without proving that $s_n=(-1)^n(2n+1)$. (user's answer is helpful.)
First, we have $$\begin{align}s_{n+1}&=\sum_{k=0}^{n+1}(-4)^k\binom{n+1+k}{2k} \\\\&=\sum_{k=0}^{n+1}(-4)^k\bigg(\binom{n+k}{2k}+\binom{n+k}{2k-1}\bigg) \\\\&=\sum_{k=0}^{\color{red}{n+1}}(-4)^k\binom{n+k}{2k}+\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1} \\\\&=\sum_{k=0}^{\color{red}n}(-4)^k\binom{n+k}{2k}+\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1} \\\\&=s_n+\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1}\end{align}$$
So, we obtain $$\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1}=s_{n+1}-s_n\tag1$$
Also, we have $$\begin{align}&\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1}\\\\&=\sum_{k=0}^{n+1}(-4)^k\bigg(\binom{n+k-1}{2k-1}+\binom{n+k-1}{2k-2}\bigg) \\\\&=\sum_{k=0}^{\color{red}{n+1}}(-4)^k\binom{n+k-1}{2k-1}+\sum_{k=\color{blue}0}^{n+1}(-4)^k\binom{n+k-1}{2k-2} \\\\&=\sum_{k=0}^{\color{red}n}(-4)^k\binom{n+k-1}{2k-1}+\sum_{k=\color{blue}1}^{n+1}(-4)^{k}\binom{n+k-1}{2k-2} \\\\&=\sum_{k=0}^{n}(-4)^k\binom{n+k-1}{2k-1}-4\sum_{k=1}^{n+1}(-4)^{k-1}\binom{n+k-1}{2k-2} \\\\&\operatorname*{=}_{\color{red}{j=k-1}}\sum_{k=0}^{n}(-4)^k\binom{n+k-1}{2k-1}-4\sum_{j=0}^{n}(-4)^{j}\binom{n+j}{2j} \\\\&=\sum_{k=0}^{n}(-4)^{k}\binom{n-1+k}{2k-1}-4s_n\end{align}$$
So, we obtain $$\sum_{k=0}^{n+1}(-4)^k\binom{n+k}{2k-1}=\sum_{k=0}^{n}(-4)^{k}\binom{n-1+k}{2k-1}-4s_n\tag2$$
Therefore, it follows from $(1)(2)$ that $$s_{n+1}-s_n=(s_{n}-s_{n-1})-4s_n$$ from which we finally get $$s_{n+1}+2s_n+s_{n-1}=0$$
