I'm getting the vibe of a progression here.

$\begin{aligned} \dfrac{f(x-1)+f(x+1)}{2f(x)}&=\Im \left(e^{\tfrac{i\pi}{3}}\right)=\dfrac{\sqrt{3}}{2}\hspace{4mm}\color{#3D99F6}\small(\text{Given}) \\\\ f(x-1)+f(x+1)&=\sqrt{3}\cdot f(x)\hspace{4mm}\color{#3D99F6}\small(1)\\ f(x)+f(x+2)&=\sqrt{3}\cdot f(x+1)\hspace{4mm}\color{#3D99F6}\small(2)\\ f(x+1)+f(x+3)&=\sqrt{3}\cdot f(x+2)\hspace{4mm}\color{#3D99F6}\small(3)\\\\ \text{Adding }\color{#3D99F6}(1) \color{#333333}\text{ and }\color{#3D99F6}(3) \color{#333333}\text{ we get,}\\\\ f(x-1)+2f(x+1)+f(x+3)&=\sqrt{3}\cdot\left(f(x)+f(x+2)\right)\\\\ &=\sqrt{3}\cdot(\sqrt{3}\cdot f(x+1))\hspace{4mm}\color{#3D99F6}\small\text{From }(2)\\\\ \implies f(x-1)+2f(x+1)+f(x+3)&=3\cdot f(x+1)\\\\ f(x-1)+f(x+3)&=f(x+1)\hspace{4mm}\color{#3D99F6}\small(4)\\ f(x+1)+f(x+5)&=f(x+3)\hspace{4mm}\color{#3D99F6}\small(5) \hspace{4mm}\text{changing x to x+2}\\\\ \text{Adding }\color{#3D99F6}(4) \color{#333333}\text{ and }\color{#3D99F6}(5) \color{#333333}\text{ we get,}\\\\ f(x-1)+f(x+5)&=0\\ f(x)+f(x+6)&=0\hspace{4mm}\color{#3D99F6}\small(6)\hspace{4mm}\text{changing x to x+1}\\ \text{Similarly,}f(x+6)+f(x+12)&=0\hspace{4mm}\color{#3D99F6}\small(7)\\\\ \text{Subtracting }\color{#3D99F6}(7) \color{#333333}\text{ from }\color{#3D99F6}(6) \color{#333333}\text{ we get,}\\\\ f(x)-f(x+12)&=0\\\\ \implies f(x)&=f(x+12)\\\\ \implies f(x+12k)&=f(x)\hspace{4mm}\color{#3D99F6}\small\text{f(x) is periodic with period 12}\\\\ \implies \sum_{k=0}^{\alpha-1}f(5+12k)&= \sum_{k=0}^{\alpha-1}f(5)\\ &=f(5)\sum_{k=0}^{\alpha-1}1\\ &=\alpha\cdot f(5)\\ &=\color{#EC7300}\boxed{\color{#333333}{\alpha}^2}\end{aligned}$