The first try.

Note that the question uses an unusual notation for combinations of k from a set of n, $\text{C}_n^k$ (as discussed in this wikipedia article ). For this solution, I will use the more common notation, $n\choose k$ .

Consider the binomial expansion of $(1+x)^{8n}$ : $(1+x)^{8n}={8n\choose0}+{8n\choose1}x+{8n\choose2}x^2+{8n\choose3}x^3+\dots+{8n\choose8n}x^{8n}$ Differentiating both sides: $8n(1+x)^{8n-1}={8n\choose1}+2{8n\choose2}x+3{8n\choose3}x^2+4{8n\choose4}x^3+\dots+8n{8n\choose8n}x^{8n-1}$ Now, let $x=i$ : $\begin{aligned}LHS&=8n(1+i)^{8n-1}\\ &=8n(\sqrt2)^{8n-1}\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)^{8n-1}\\ &=8\sqrt2n(\sqrt2)^{8n-2}\left[\cos\left(\frac\pi4\right)+i\sin\left(\frac\pi4\right)\right]^{8n-1}\\ &=2^3\sqrt2n\times2^{4n-1}\left[\cos\left(\frac\pi4(8n-1)\right)+i\sin\left(\frac\pi4(8n-1)\right)\right]&\text{by de Moivre's theorem}\\ &=\sqrt2n\times2^{4n+2}\left[\cos\left(2\pi n-\frac\pi4\right)+i\sin\left(2\pi n-\frac\pi4\right)\right]\\ &=\sqrt2n\times2^{4n+2}\left[\cos\left(-\frac\pi4\right)+i\sin\left(-\frac\pi4\right)\right]\\ &=\sqrt2n\times2^{4n+2}\left(\frac1{\sqrt2}-\frac i{\sqrt2}\right)\\ &=2^{4n+2}(1-i)n\end{aligned}$ $\begin{aligned}RHS&={8n\choose1}+2{8n\choose2}i+3{8n\choose3}i^2+4{8n\choose4}i^3+5{8n\choose5}i^4+\dots+8n{8n\choose8n}i^{8n-1}\\ &={8n\choose1}+2{8n\choose2}i-3{8n\choose3}-4{8n\choose4}i+5{8n\choose5}+\dots+8n{8n\choose8n}i^{8n-1}\\ &=\left[{8n\choose1}-3{8n\choose3}+5{8n\choose5}+\dots-(8n-1){8n\choose8n-1}\right]+\left[2{8n\choose2}-4{8n\choose4}+6{8n\choose6}+\dots-8n{8n\choose8n}\right]i\end{aligned}$ Equating real parts: $\begin{aligned}2^{4n+2}n&={8n\choose1}-3{8n\choose3}+5{8n\choose5}+\dots-(8n-1){8n\choose8n-1}\\ \therefore2^{4n+2}n&=n^{10}\\ \therefore4^{2n+1}&=n^9\end{aligned}$ By inspection, this is true when $\boxed{n=4}$