In given series difference of suffies is 4
$$
0-4=8-4=12-4=\ldots=4
$$

Now, $(\mathrm{l})^{\frac{1}{4}}=(\cos 0+i \sin 0)^{\frac{1}{4}}$
$$
\begin{aligned}
& =(\cos 2 r \pi+i \sin 2 r \pi)^{\frac{1}{4}} \\
& =\cos \frac{r \pi}{2}+i \sin \frac{r \pi}{2}
\end{aligned}
$$
where $r=0,1,2,3$
Four roots of unity are $1, i,-1,-i$
$$
\begin{aligned}
& =1, \alpha, \alpha, \alpha^2 \quad \text{say}\\
\text { and }(1+x)^n & =\sum_{r=0}^n{ }^n C_r x^r
\end{aligned}
$$


For $r=0,4,8,12 \ldots$ R.H.S of Eq. (v)
$$
\begin{array}{r}
{ }^n C_0(1+1+1+1)+{ }^n C_4\left(1+\alpha^4+\alpha^8+\alpha^{12}\right) \\
+{ }^n C_8\left(1+\alpha^8+\alpha^{16}+\alpha^{24}\right)+\ldots+ \\
=4\left({ }^n C_0+{ }^n C_4+{ }^n C_8 \ldots\right) \quad\left(\because \alpha^4=1\right) .
\end{array}
$$
L.H.S of Eq. (v)
$$
\begin{aligned}
& =2^n+(1+i)^n+\left(1+i^2\right)^n+\left(1+i^3\right)^n \\
& =2^n+(1+i)^n+0+(1-i)^n
\end{aligned}
$$

Since, $(1+i)^n=2^{\frac{n}{2}}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^n$
$$
=2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right]
$$
$$
\begin{aligned}
& \quad(1-i)^n=2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}-i \sin \frac{n \pi}{4}\right] \\
& \Rightarrow 2^n+(1+i)^n+(1-i)^n \\
& =2^n+2 \cdot 2^{\frac{n}{2}} \cos \frac{n \pi}{4} \\
& \Rightarrow 4 \cdot\left({ }^n C_0+{ }^n C_4+{ }^n C_8+{ }^n C_{12} \cdots\right) \\
& \quad=2 \cdot 2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]=\frac{2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+2^{\frac{n}{2}-1}\right]}{2}
\end{aligned}
$$