Prove: ${n\choose 1}-3{n\choose 3}+9{n\choose 5}-...=\frac{-1}{\sqrt{3}}(-2)^n\sin\frac{2n\pi}{3}$ Prove: ${n\choose 1}-3{n\choose 3}+9{n\choose 5}-...=\frac{-1}{\sqrt{3}}(-2)^n\sin\frac{2n\pi}{3}$
How to use binomial theorem on a left sum?
 A: We could also start with $\sin\left(\frac{2\pi n}{3}\right)$ and use de Moivre's formula
in order to derive the binomial expression on the LHS.

The following is valid
  \begin{align*}
\sum_{k}\binom{n}{2k+1}(-1)^k3^k=-\frac{1}{\sqrt{3}}(-2)^n\sin\left(\frac{2\pi n}{3}\right)\qquad n>0\\
\end{align*}

$$ $$

We obtain for $n>0$
  \begin{align*}
\sin\left(\frac{2\pi n}{3}\right)&=\Im\left(\cos\left(\frac{2\pi n}{3}\right)+i\sin\left(\frac{2\pi n}{3}\right)\right)\tag{1}\\
&=\Im\left(\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right)^n\\
&=\Im\left(\sum_{k=0}^{n}\binom{n}{k}i^k\sin^k\left(\frac{2\pi}{3}\right)\cos^{n-k}\left(\frac{2\pi}{3}\right)\right)\\
&=\sum_{k}\binom{n}{2k+1}(-1)^k\sin^{2k+1}\left(\frac{2\pi}{3}\right)\cos^{n-(2k+1)}\left(\frac{2\pi}{3}\right)\tag{2}\\
&=\sum_{k}\binom{n}{2k+1}(-1)^k\left(\frac{\sqrt{3}}{2}\right)^{2k+1}\left(-\frac{1}{2}\right)^{n-(2k+1)}\\
&=-\sqrt{3}\left(-\frac{1}{2}\right)^n\sum_{k}\binom{n}{2k+1}(-1)^k3^k\\
\end{align*}
and the claim follows.

Comment:


*

*In (1) we apply de Moivre's formula.

*In (2) we take odd index values $2k+1$ since we need the imaginary part only.
A: Hint: consider the binomial expansions of $(1+ix)^n$ and $(1-ix)^n$, their sum and their difference.
A: Hint:
Binomial expansion of $\;\Bigl(\dfrac{1+\mathrm i\sqrt 3}2\Bigr)^n=\mathrm e^{\tfrac{2n\mathrm i\pi}3}$.
