Show that $\binom{n}{1}-3\binom{n}{3}+3^2\binom{n}{5}\cdots=0$

Question

Show that if $n\equiv 0\pmod 6$ (although the statement holds true for $n\equiv 0\pmod 3$)

$\binom{n}{1}-3\binom{n}{3}+3^2\binom{n}{5}\cdots=0$

I am having trouble finding the appropriate polynomial to resolve this sum. Any hints? I prefer hints to complete solutions. I also have tried to come up with a probability story that gives me the relation above to no avail.

Answer 1

By virtue of the binomial series, we have that $$\binom{n}{1} - 3 \binom{n}{3} + 3^2 \binom{n}{5} + \ldots = \sum_{k=0}^\infty \binom{n}{2k+1} (-3)^k = \frac{2^n}{\sqrt{3}} \sin\left(\frac{n\pi}{3}\right).$$ Since $n \equiv 0 \pmod{6}$, we have that $n=6k$ for some $k\in\mathbb{Z}$. Plug that in and you will have the result.

Answer 2

Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$ we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$ Now take $a_k=\binom{n}{k+1}(i\sqrt{3})^k$ to obtain \begin{align} \sum_{k\ge 0} \binom{n}{2k+1}(-3)^k &=\sum_{k\ge 0} \binom{n}{2k+1}(i\sqrt{3})^{2k} \\ &= \sum_{k\ge 0} \binom{n}{k+1}(i\sqrt{3})^k \frac{1+(-1)^k}{2} \\ &= \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1}(i\sqrt{3})^k + \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1}(-i\sqrt{3})^k \\ &= \frac{1}{2i\sqrt{3}}\sum_{k\ge 0} \binom{n}{k+1}(i\sqrt{3})^{k+1} - \frac{1}{2i\sqrt{3}}\sum_{k\ge 0} \binom{n}{k+1}(-i\sqrt{3})^{k+1} \\ &= \frac{1}{2i\sqrt{3}}\sum_{k\ge 1} \binom{n}{k}(i\sqrt{3})^k - \frac{1}{2i\sqrt{3}}\sum_{k\ge 1} \binom{n}{k}(-i\sqrt{3})^k \\ &= \frac{1}{2i\sqrt{3}}\left((1+i\sqrt{3})^n - 1\right) - \frac{1}{2i\sqrt{3}}\left((1-i\sqrt{3})^n-1\right) \\ &= \frac{(1+i\sqrt{3})^n-(1-i\sqrt{3})^n}{2i\sqrt{3}} \\ &= \frac{(\cos(\pi/3)+i\sin(\pi/3))^n-(\cos(\pi/3)-i\sin(\pi/3))^n}{2i\sqrt{3}} \\ &= \frac{2^n(\cos(n\pi/3)+i\sin(n\pi/3))-2^n(\cos(n\pi/3)-i\sin(n\pi/3))}{2i\sqrt{3}} \\ &= \frac{2^n i\sin(n\pi/3)+2^n i\sin(n\pi/3)}{2i\sqrt{3}} \\ &= \frac{2^n\sin(n\pi/3)}{\sqrt{3}} \end{align}