$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=$

Question

If $n=12m$, where $m\in\Bbb{N}$, prove that $$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=(-1)^m\left(\frac{2\sqrt2}{1+\sqrt3}\right)^n$$

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I tried to solve it but stuck after few steps:

$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....$
$=1-\frac{1}{(2+\sqrt3)^2}\frac{n(n-1)}{2}+\frac{1}{(2+\sqrt3)^4}\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}+....$

How should I prove this question?

Answer 1

Hint: First try to prove that the sum can be simplified to $$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=\operatorname{Re}\left(\frac{1}{2+\sqrt3}+i\right)^n.$$

Answer 2

Using $\displaystyle (1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+............$

Now Put $\displaystyle x=\frac{i}{2+\sqrt{3}}$ and $\displaystyle x=-\frac{i}{2+\sqrt{3}}$ Respectively, We get

$\displaystyle \left(1+\frac{i}{2+\sqrt{3}}\right)^{n} = \binom{n}{0}+\binom{n}{1}\cdot \frac{i}{2+\sqrt{3}}+\binom{n}{2}\cdot \frac{i^2}{(2+\sqrt{3})^2}+............(1)$

$\displaystyle \left(1-\frac{i}{2+\sqrt{3}}\right)^{n} = \binom{n}{0}-\binom{n}{1}\cdot \frac{i}{2+\sqrt{3}}+\binom{n}{2}\cdot \frac{i^2}{(2+\sqrt{3})^2}+...........(2)$

Now Add these two equations, We get

$\displaystyle \frac{[1+i(2-\sqrt{3})]^n+[1-i(2-\sqrt{3})]^n}{2} =\binom{n}{0}-\binom{n}{2}\cdot \frac{1}{(2+\sqrt{3})^2}+\binom{n}{4}\cdot \frac{1}{(2+\sqrt{3})^4}$