How much Strong is your Binomial!!!

Relevant wiki: Binomial Theorem - Expansions - Basic

Let the sum be $\displaystyle S= \sum_{k=0}^{n/2} \frac {(-1)^k \binom n{2k}}{(2-\sqrt 3)^{2k}}$ . Using binomial theorem, we have:

$\begin{aligned} (1+z)^n & = \sum_{k=0}^n \binom nk z^k & \small \color{#3D99F6} \text{where }n=12m \\ (1-z)^n & = \sum_{k=0}^n (-1)^k \binom nk z^k \\ \frac {(1+z)^n+(1-z)^n}2 & = \sum_{k=0}^{n/2} \binom n{2k} z^{2k} & \small \color{#3D99F6} \text{Let }z = ix \text{, where }i = \sqrt{-1} \\ & = \sum_{k=0}^{n/2} (-1)^k \binom n{2k} x^{2k} & \small \color{#3D99F6} \text{Let }x = \frac 1{2-\sqrt 3} \\ & = \sum_{k=0}^{n/2} \frac {(-1)^k \binom n{2k}}{(2-\sqrt 3)^{2k}} \end{aligned}$

$\begin{aligned} \implies S & = \frac {(1+z)^n+(1-z)^n}2 & \small \color{#3D99F6} \text{where }z = \frac i{2-\sqrt 3} = (2+\sqrt 3)i \\ & = \frac {(1+(2+\sqrt 3)i)^n+(1-(2+\sqrt 3)i)^n}2 \\ & = \frac 12 \left(\sqrt{8+4\sqrt 3}\right)^n \left(e^{n \tan^{-1} (2+\sqrt 3) i} + e^{-n \tan^{-1} (2+\sqrt 3)i}\right) & \small \color{#3D99F6} \text{Note that } \tan^{-1} (2+\sqrt 3) = \frac {5\pi}{12} \\ & = \left(\sqrt{8+4\sqrt 3}\right)^n \cos (5m\pi) \\ & = (-1)^m \left(\sqrt {2(\sqrt 3+1)^2} \right)^n \\ & = (-1)^m \left(\sqrt 2 (\sqrt 3+1) \right)^n \\ & = (-1)^m \left(\frac {\sqrt 8}{\sqrt 3-1} \right)^n \end{aligned}$

Therefore, $\dbinom TS + T^S = \dbinom 83 + 8^3 = 56+512 = \boxed{568}$ .

T = 8, S =3 therefore $^T C_S + T^S$ = $^8 C_3 + 8^3$ = 56 +512 = 568