If n is a positive multiple of 6, show that
$${n\choose1}-3{n\choose3}+3^2{n\choose5}-\cdots=0$$
$${n\choose1}-\frac{1}{3}{n\choose3}+\frac{1}{3^2}{n\choose5}-\cdots=0$$
I know that $\frac{1}{2}[(1+x)^n-(1-x)^n]$ will isolate the odd terms in the expansion, but I'm not sure how to make the series alternate and have the corresponding powers.
As you already stated,
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$$
and
\begin{align*} \frac{1}{2}\left((1+x)^n - (1-x)^n\right) &= \sum_{k \geq 0} \binom{n}{2k+1} x^{2k+1}\\ &= \binom{n}{1}x + \binom{n}{3}x^3 + \binom{n}{5}x^5 + \ldots \end{align*}
Now, to get alternate powers of odd terms, what rings a bell, is that $i^{4k+1} = i$ and $i^{4k+3} = -i$. So we use, this to get the required result as
\begin{align*} \frac{1}{2}\left((1+ix)^n - (1-ix)^n\right) &= \binom{n}{1}ix + \binom{n}{3}(ix)^3 + \binom{n}{5}(ix)^5 + \ldots \\ &= i\left(\binom{n}{1}x - \binom{n}{3}x^3 + \binom{n}{5}x^5 + \ldots\right) \\ \end{align*}
Thus,
$$\sum_{k\geq 0} \binom{n}{2k+1} (-1)^k x^{2k} = \frac{i}{2x}\left((1-ix)^n - (1+ix)^n\right)$$
Substituting $x=\sqrt(3)$ yields
$$\left(\binom{n}{1} - \binom{n}{3}3 + \binom{n}{5}3^2 + \ldots\right) = \frac{i}{2\sqrt{3}} ((1-\sqrt(3)i)^n - (1+\sqrt(3)i)^n) = \frac{i}{2} (e^{-in\pi/3}-e^{in\pi/3})$$
Since, $n = 6m$, on RHS we get $\frac{i}{2\sqrt{3}} (e^{-2im\pi} - e^{2im\pi}) = 0$
Similarly, substitute $x=\frac{1}{\sqrt{3}}$ to get the next result.
A related way to look at is is as follows.
The expressions are the imaginary part of $\frac{1}{x}(1+xi)^n$ for $x=\sqrt{3}$ and $x=\frac{1}{\sqrt{3}}$. ] But $1+\sqrt{3}i = 2e^{i\pi/3}$ and $1+\frac{1}{\sqrt{3}}i = \frac{2}{\sqrt{3}}e^{i\pi/6}$.
So you actually get that the values are $2^{n}\sin \frac{n\pi}{3}$ for any $n$ in the first case, and the value $\left(\frac{2}{\sqrt{3}}\right)^n\sin\frac{n\pi}{6}.$