Prove: ${n\choose 1}-3{n\choose 3}+9{n\choose 5}-...=\frac{-1}{\sqrt{3}}(-2)^n\sin\frac{2n\pi}{3}$

Question

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?

Answer 1

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*}

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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.

Answer 2

Hint: consider the binomial expansions of $(1+ix)^n$ and $(1-ix)^n$, their sum and their difference.

Answer 3

Hint:

Binomial expansion of $\;\Bigl(\dfrac{1+\mathrm i\sqrt 3}2\Bigr)^n=\mathrm e^{\tfrac{2n\mathrm i\pi}3}$.