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Evaluate $$\sum ^n _{r=0} \binom{n}{r} \tan^{2r}\left(\frac \pi 3 \right)$$

So I've got to a point at which I don't know how to go any further, any help would be appreciated. My workings so far are shown. $$\sum ^n _{r=0} \frac {n!}{r!(n-r)!}\tan^{2r}\left(\frac \pi 3 \right)$$ $$n! \sum ^n _{r=0} \frac {1}{r!(n-r)!}\tan^{2r}\left(\frac \pi 3 \right)$$ $$n! \sum ^n _{r=0} \frac {1}{r!(n-r)!} \times(\sqrt 3)^{2r}$$ $$n! \sum ^n _{r=0} \frac {3^r}{r!(n-r)!}$$

This is as far as I've been able to get, any help would be appreciated.

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    $\begingroup$ looks like a job for the binomial theorem! $\endgroup$ Feb 11 '19 at 18:11
  • $\begingroup$ binomial theorem indeed since $\tan^{2r}\left(\frac \pi 3 \right)=\left(\tan^{2}\left(\frac \pi 3 \right)\right)^r$. $\endgroup$
    – rtybase
    Feb 11 '19 at 18:13
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As already mentioned by Lord Shark the Unknown we may use the Binomial Theorem here. Note that

$$\sum_{r=0}^n \binom nr \tan^{2r}\left(\frac\pi3\right)=\sum_{r=0}^n \binom nr (1)^{n-r}\left(\tan^2\left(\frac\pi3\right)\right)^r=\left(1+\tan^2\left(\frac\pi3\right)\right)^n$$

Utilizing the fact that $1+\tan^2(x)=\sec^2(x)$ aswell as the well-known $\cos\left(\frac\pi3\right)=\frac12$ we obtain that

$$\left(1+\tan^2\left(\frac\pi3\right)\right)^n=\left(\sec^2\left(\frac\pi3\right)\right)^n=\left(\frac1{\cos^2\left(\frac\pi3\right)}\right)^n=(2^2)^n$$

$$\therefore~\sum_{r=0}^n \binom nr \tan^{2r}\left(\frac\pi3\right)~=~4^n$$

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  • $\begingroup$ I'm folliwng everything in you answer except the very first line. Why is this the case. $$\sum_{r=0}^n \binom nr \tan^{2r}\left(\frac\pi3\right)=\sum_{r=0}^n \binom nr (1)^{n-r}\left(\tan^2\left(\frac\pi3\right)\right)^r$$. $\endgroup$
    – H.Linkhorn
    Feb 11 '19 at 19:14
  • $\begingroup$ @H.Linkhorn In general we know due the Binomial Theorem that $$(a+b)^n=\sum_{r=0}^n\binom nr a^{n-r}b^r$$ Now my aim was to show that the given structure can be identified as such. Therefore I firstly used that $a^{bc}=(a^b)^c$ to rewrite the tangent term and since $1^{n-r}=1$ this does not change the outcome either but we got the clearly recognizable structure of $a^{n-r}b^r$. Everything clear now? $\endgroup$
    – mrtaurho
    Feb 11 '19 at 19:18
  • $\begingroup$ Yeah thats great. Thank you $\endgroup$
    – H.Linkhorn
    Feb 11 '19 at 19:19
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hint

$$(1+t)^n=\sum_{r=0}^n\frac{n!}{r!(n-r)!}t^r$$

use $$1+\tan^2(x)=\frac{1}{\cos^2(x)}$$

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The final form will be $\sum_{r=0}^{n}\displaystyle\frac{3^rn!}{(n-r)!r!} = (1+3)^n = 4^n$

Recall that $\sum_{r=0}^{n}$ ${n}\choose{r}$ ${x^r} = (1+x)^n$

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