已知 $f(x)$ 和 $g(x)$ 的定义域均为 $\mathbb R$，且满足\[\begin{cases} f(x+1)=-\dfrac 12f(x)+\dfrac{\sqrt 3}2g(x),\\ g(x+1)=-\dfrac 12g(x)-\dfrac{\sqrt 3}2f(x),\end{cases}\]且 $f(x)=f(5-x)$，$g(365)=-\sqrt 3$，则 $\displaystyle\sum_{k=1}^{2023}f(k)=$_______．
答案 $2$．
解析 根据题意，有\[g(x)=\dfrac{2}{\sqrt 3}f(x+1)+\dfrac{1}{\sqrt 3}f(x),\]于是\[\left(\dfrac{2}{\sqrt 3}f(x+1)+\dfrac{1}{\sqrt 3}f(x+1)\right)=-\dfrac 12\left(\dfrac{2}{\sqrt 3}f(x+1)+\dfrac{1}{\sqrt 3}f(x)\right)-\dfrac{\sqrt 3}2f(x),\]整理得\[f(x+2)=-f(x+1)-f(x),\]设 $f(1)=a$，$f(2)=b$，则\[f(k):\underbrace{a,b,-a-b},\underbrace{a,b,-a-b},\cdots,\]进而\[\begin{cases} f(x)=f(5-x),\\ g(365)=-\sqrt 3,\end{cases}\implies \begin{cases} f(2)=f(3),\\ 2f(366)+f(365)=-3,\end{cases}\implies \begin{cases} b=-a-b,\\ 2(-a-b)+b=-3,\end{cases}\implies \begin{cases} a=2,\\ b=-1,\end{cases}\]因此 $\displaystyle\sum_{k=1}^{2023}f(k)=f(1)=a=2$．
$\left(\frac{2}{\sqrt{3}} f(x+2)+\frac{1}{\sqrt{3}} f(x+1)\right)=-\frac{1}{2}\left(\frac{2}{\sqrt{3}} f(x+1)+\frac{1}{\sqrt{3}} f(x)\right)-\frac{\sqrt{3}}{2} f(x)$