You may need help of complex numbers

Let $i = \sqrt{-1}$ . Recall that the binomial theorem says, for any complex $x$ and non-negative integer $n$ ,

$\displaystyle (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$

By the convention that $\binom{n}{k} = 0$ when $k > n$ , we can change the upper bound:

$\displaystyle (1+x)^n = \sum_{k=0}^\infty \binom{n}{k} x^k$

And now we can break it into four separate series according to the value of $k \mod 4$ :

$\displaystyle (1+x)^n = \sum_{k=0}^\infty \binom{n}{4k} x^{4k} + \sum_{k=0}^\infty \binom{n}{4k+1} x^{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} x^{4k+2} + \sum_{k=0}^\infty \binom{n}{4k+3} x^{4k+3}$

Substituting $x = 1$ ,

$\displaystyle\begin{aligned} (1+1)^n &= \sum_{k=0}^\infty \binom{n}{4k} 1^{4k} + \sum_{k=0}^\infty \binom{n}{4k+1} 1^{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} 1^{4k+2} + \sum_{k=0}^\infty \binom{n}{4k+3} 1^{4k+3} \\ &= \sum_{k=0}^\infty \binom{n}{4k} + \sum_{k=0}^\infty \binom{n}{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} + \sum_{k=0}^\infty \binom{n}{4k+3} \end{aligned}$

Similarly, we can do the same for $x = i, -1, -i$ :

$\displaystyle\begin{aligned} (1+i)^n &= \sum_{k=0}^\infty \binom{n}{4k} i^{4k} + \sum_{k=0}^\infty \binom{n}{4k+1} i^{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} i^{4k+2} + \sum_{k=0}^\infty \binom{n}{4k+3} i^{4k+3} \\ &= \sum_{k=0}^\infty \binom{n}{4k} + i \sum_{k=0}^\infty \binom{n}{4k+1} - \sum_{k=0}^\infty \binom{n}{4k+2} - i \sum_{k=0}^\infty \binom{n}{4k+3} \\ (1-1)^n &= \sum_{k=0}^\infty \binom{n}{4k} (-1)^{4k} + \sum_{k=0}^\infty \binom{n}{4k+1} (-1)^{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} (-1)^{4k+2} + \sum_{k=0}^\infty \binom{n}{4k+3} (-1)^{4k+3} \\ &= \sum_{k=0}^\infty \binom{n}{4k} - \sum_{k=0}^\infty \binom{n}{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} - \sum_{k=0}^\infty \binom{n}{4k+3} \\ (1-i)^n &= \sum_{k=0}^\infty \binom{n}{4k} (-i)^{4k} + \sum_{k=0}^\infty \binom{n}{4k+1} (-i)^{4k+1} + \sum_{k=0}^\infty \binom{n}{4k+2} (-i)^{4k+2} + \sum_{k=0}^\infty \binom{n}{4k+3} (-i)^{4k+3} \\ &= \sum_{k=0}^\infty \binom{n}{4k} - i \sum_{k=0}^\infty \binom{n}{4k+1} - \sum_{k=0}^\infty \binom{n}{4k+2} + i \sum_{k=0}^\infty \binom{n}{4k+3} \end{aligned}$

Now, observe that $(1+1)^n + i(1+i)^n - (1-1)^n - i(1-i)^n$ add up to exactly $4 \sum_{k=0}^\infty \binom{n}{4k+3}$ . (The trick is that we convert the coefficients so that they become the sum of roots of unity, except for the $\sum \binom{n}{4k+3}$ one which now becomes all $1$ 's.)

Additionally, let $E(\theta) = \cos \theta + i \sin \theta$ . ( $E$ is more known as $\text{cis}$ , but we also have $E(\theta) = e^{i \theta}$ . Also $\text{cis}$ is much harder to type in LaTeX.) Note that $E(\theta)^n = E(n\theta)$ . Note that $(1+i)^n = \left( \sqrt{2} E \left( \frac{\pi}{4} \right) \right)^n = 2^{n/2} E \left( \frac{n\pi}{4} \right)$ , and likewise $(1-i)^n = \left( \sqrt{2} E \left( \frac{-\pi}{4} \right) \right)^n = 2^{n/2} E \left( \frac{-n\pi}{4} \right)$ . Thus,

$\displaystyle\begin{aligned} \sum_{k=0}^\infty \binom{n}{4k+3} &= \frac{1}{4} ((1+1)^n + i(1+i)^n - (1-1)^n - i(1-i)^n) \\ &= \frac{1}{4} \left( 2^n + i \cdot 2^{n/2} E \left( \frac{n\pi}{4} \right) - 0 - i \cdot 2^{n/2} E \left( \frac{-n\pi}{4} \right) \right) \\ &= \frac{1}{4} \left( 2^n + i \cdot 2^{n/2} \cdot \left( E \left( \frac{n\pi}{4} \right) - E \left( \frac{-n\pi}{4} \right) \right) \right) \\ &= \frac{1}{4} \left( 2^n + i \cdot 2^{n/2} \cdot \left( \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) - \left( \cos \frac{-n\pi}{4} + i \sin \frac{-n\pi}{4} \right) \right) \right) \\ &= \frac{1}{4} \left( 2^n + i \cdot 2^{n/2} \cdot \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} - \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) \right) \\ &= \frac{1}{4} \left( 2^n + i \cdot 2^{n/2} \cdot 2i \cdot \sin \frac{n\pi}{4} \right) \\ &= \boxed{ \frac{1}{2} \left( 2^{n-1} - 2^{n/2} \sin \frac{n\pi}{4} \right) } \end{aligned}$

Made typical use of induction to eliminate option using n=3 &n=4 which leads to the correct option within a minute!