Binomial + Trigo + Complex!

We know that, for even $n$ ,

$\begin{aligned} (1+xi)^n & = ^nC_0 + ^n C_1 xi - ^nC_2 x^2 - ^nC_3 x^3i + ^nC_4 x^4 + ... +(-1)^{\frac n2 \ n}C_n x^n & \small \color{#3D99F6}{\text{Taking the real part}} \\ \Re \left \{(1+xi)^n \right \} & = ^nC_0 - ^nC_2 x^2 + ^nC_4 x^4 - ^nC_6 x^6 + ... +(-1)^{\frac n2 \ n}C_n x^n & \small \color{#3D99F6}{\text{Putting } x= \frac 1{2+\sqrt 3}} \\ \Re \left \{\left(1+\frac i{2+\sqrt 3}\right)^n \right \} & = ^nC_0 - \frac {^nC_2}{(2+\sqrt 3)^2} + \frac {^nC_4}{(2+\sqrt 3)^4} - \frac {^nC_6}{(2+\sqrt 3)^6} + ... +\frac {(-1)^{\frac n2 \ n}C_n}{(2+\sqrt 3)^n} \end{aligned}$

Therefore, the sum is given by:

$\begin{aligned} S & = \Re \left \{\left(1+\frac i{2+\sqrt 3}\right)^n \right \} \\ & = \Re \left \{\left(\sqrt{1+\frac 1{(2+\sqrt 3)^2}}\right)^n e^{n \tan^{-1} \frac 1{2+\sqrt 3}i} \right \} \\ & = \Re \left \{\left(\sqrt{8-4\sqrt 3}\right)^n e^{\frac {n \pi}{12}i} \right \} \\ & = \Re \left \{\left(\frac {\sqrt 8}{1+\sqrt 3} \right)^n \left(\cos \frac {n \pi}{12} + i \sin \frac {n \pi}{12} \right) \right \} \\ & = \left(\frac {\sqrt 8}{1+\sqrt 3} \right)^n \cos \frac {n \pi}{12} \\ & = \cos (m \pi) \left(\frac {\sqrt 8}{1+\sqrt 3} \right)^n \\ & = (-1)^m \left(\frac {\sqrt 8}{1+\sqrt 3} \right)^n \end{aligned}$

$\implies P-Q = 8-3 = \boxed{5}$

First we can notice that $\tan \frac{pi}{12}=\frac{\sin \frac{pi}{12}}{\cos\frac{pi}{12}}=\frac{\sin \frac{pi}{12}*2\cos\frac{pi}{12}}{2\cos^{2}\frac{pi}{12}}=\frac{\sin \frac{\pi}{6}}{1+\cos\frac{\pi}{6}}=\frac{\frac{1}{2}}{1+\frac{\sqrt 3}{2}}=\frac{1}{2+\sqrt 3}$ . Using that $\frac{\sin \frac{\pi}{12}}{\cos\frac{\pi}{12}}=\frac{1}{2+\sqrt 3}$ and $\sin^2 \frac{\pi}{12}+\cos^2 \frac{\pi}{12}=1,$ we can also verify that $\cos \frac{\pi}{12}= \frac{1+\sqrt 3 }{2\sqrt 2 }$ and $\sin \frac{\pi}{12}= \frac{\sqrt {3} -1}{2\sqrt 2}.$

Assuming that $n=12m$ , where $m$ is any positive integer, we have the following

$\sum_{k=0}^{k=\frac{n}{2}}(-1)^{k}{n\choose 2 k} \frac{1}{(2+\sqrt 3 )^{2*k}}=\sum_{k=0}^{k=\frac{n}{2}}(-1)^{k}{n\choose 2k} \tan^{2k}\frac{\pi}{12}$ $=\frac{1}{2}((1+i \tan \frac{\pi}{12})^{n}+(1-i \tan \frac{\pi}{12})^{n})$ $=\frac{1}{2\cos^n \frac{\pi}{12}}((\cos \frac{\pi}{12}+i \sin \frac{\pi}{12})^{n}+(\cos\frac{\pi}{12}-i \sin \frac{\pi}{12})^{n})$ $=\frac{1}{2\cos^n \frac{\pi}{12}}((\cos \frac{n \pi}{12}+i \sin \frac{n\pi}{12})+(\cos \frac{n\pi}{12}-i \sin \frac{n\pi}{12}))$ $=\frac{1}{2\cos^n \frac{\pi}{12}}(2\cos \frac{n \pi}{12})=\frac{1}{\cos^n \frac{\pi}{12}}(\cos {m \pi})=\frac{1}{\cos^n \frac{\pi}{12}}(-1)^{m}$ $=(-1)^{m} (\frac{\sqrt{8}}{1+\sqrt{3}})^{n}.$

Therefore $P - Q=8 -3 =5.$

Well, easy peasy. All what you need is observation.

The sum is just the same as - $\displaystyle \sum_{k=0}^{n/2}{{n \choose 2k}\frac{{(-1)}^{k}}{{(2+\sqrt{3})}^{2k}}}$

$\displaystyle = \sum_{k=0}^{n/2}{{n \choose 2k}\frac{{(i)}^{2k}}{{(2+\sqrt{3})}^{2k}}}$

So, it must be

$\displaystyle \text{Re}({(1 + \frac{\iota}{2+\sqrt{3}})}^{n})$

$\displaystyle \text{Re}({(1 + \iota(2-\sqrt{3}))}^{n})$

When we change into polar form, we get -

$\displaystyle {(\sqrt{6}-\sqrt{2})({e}^{\pi/12\iota})}^{n}$

$\displaystyle = {(\sqrt{6}-\sqrt{2})}^{n}({e}^{\pi m \iota})$

$\displaystyle = {(-1)}^{m}({\frac{2\sqrt{2}}{1+\sqrt{3}}}^{n})$