Another nested radical

$\begin{aligned} a & = \sqrt { 2\cdot 8^{ 2 }+\sqrt { 2\cdot 8^{ 4 }+\sqrt { 2\cdot 8^{ 8 }+\cdots} } } \\ 8a & = \sqrt { 2\cdot 8^{ 4 }+\sqrt { 2\cdot 8^{ 8 }+\sqrt { 2\cdot 8^{ 16 }+\cdots} } } \\ & = a^2-2\cdot 8^2 \\ 0 & = a^2-8a-128 \\ a & =\frac{8+\sqrt{8^2-4(1)(-128)}}{2(1)} \\ & = 16 \\ \Rightarrow \frac{a}{2} & = \boxed{8} \end{aligned}$

Let's Generalise this:( for a positive integer k) $\large\mathcal{S}= \sqrt{ 2k^2+ \sqrt{ 2k^4 + \sqrt{2k^8 + \ldots } } }$ Let's multiply both sides by $\color{#D61F06}{k}$ $\color{#D61F06}{k}\times\mathcal{S}= \sqrt{\color{#D61F06}{k^2}\times2k^2+ \sqrt{ \color{#D61F06}{k^4}\times2k^4+ \sqrt{\color{#D61F06}{k^8}\times2k^8 + \ldots } } }$ $=\sqrt{ 2k^4+ \sqrt{ 2k^8 + \sqrt{2k^{16} + \ldots } } } =\mathcal{S}^2-2k^2$ $\Large\Rightarrow \mathcal{S}^2 -k\mathcal{S} -2k^2=0$ $\Large \Rightarrow ( \mathcal{S}+k)( \mathcal{S}-2k)=0$ Since $\mathcal{S}$ cannot be negative, hence $\mathcal{S}$ =2k=a.
Hence $\dfrac{a}{2}=k$ . In the given question k=8. $\huge \therefore \dfrac{a}{2}=\boxed{\color{#007fff}{8}}$