Question

If $n=12m(m\in N),{}^n{C}_0-\frac{{}^n{C}_2}{(2+\sqrt{3}{)}^2}+\frac{{}^n{C}_4}{(2+\sqrt{3}{)}^4}-\frac{{}^n{C}_6}{(2+\sqrt{3}{)}^6}+..=(-1{)}^m{\left(\frac{\sqrt{p}}{1+\sqrt{Q}}\right)}^n$ then $P-Q$ is:

Answer 1

$\frac{{(1+x)}^n+{(1-x)}^n}{2}={}^n{C}_0+{{}^{n}C}_2{x}^2+.....$

Putting $x=\frac{i}{2+\sqrt{3}}$

$\Rightarrow S=\frac{{(1+\frac{i}{2+\sqrt{3}})}^n+{(1-\frac{i}{2+\sqrt{3}})}^n}{2}$

$={{}^{n}C}_0-{{}^{n}C}_2.\frac{1}{{(2+\sqrt{3})}^2}+.....$

$\Rightarrow S=\frac{{(2+\sqrt{3}+i)}^n+{(2+\sqrt{3}+i)}^n}{2{(2+\sqrt{3})}^n}$

$=\frac{2^n[{(1+\frac{\sqrt{3}+i}{2})}^n+(1+{\left(\frac{\sqrt{3}-i}{2}\right)}^n)]}{2.2^n{(1+\sqrt{3/4})}^n}$

$=\frac{1}{2{(1+\sqrt{3/4})}^n}[{(1+e^{i\pi /6})}^n+{(1+e^{-i\pi /6})}^n]$

$=\frac{1}{2}\times \frac{1}{{(1+\frac{\sqrt{3}}{\sqrt{4}})}^n}\times 2$

$=\frac{1}{(1+\frac{\sqrt{3}}{\sqrt{4}})}$

$P=1,\theta =\frac{3}{4}$

$\therefore P-Q=\frac{1}{4}$