does anyone have solutions?

We know that $\binom{n}{1}x+\binom{n}{3}x^3+\binom{n}{5}x^5+\dotsc=\frac{1}{2} \left( (1+x)^n-(1-x)^n \right)$ from the Binomial Theorem. Dividing the whole sum by $x$ and then inserting $x=\sqrt{-\frac{1}{3}}$, we obtain that the desired sum is equal to \[\frac{i\sqrt{3}}{2} \cdot \left( \left(1-\frac{i}{\sqrt{3}} \right)^n-\left(1+\frac{i}{\sqrt{3}} \right)^n \right)=\boxed{2^n \cdot 3^{\frac{1-n}{2}} \cdot \sin \frac{n\pi}{6}}\]

\begin{align*} \sum_{k=0}^\infty\frac{1}{(-3)^k}\binom{n}{2k+1}&=\Re\Bigg(-i\sqrt{3}\sum_{k=0}^n\binom{n}{k}\bigg(\frac{i}{\sqrt{3}}\bigg)^k\Bigg)\\ &=\Re\Bigg(-i\sqrt{3}\bigg(\frac{\sqrt{3}+i}{\sqrt{3}}\bigg)^n\Bigg)\\ &=\Re(-i2^n3^{-n/2}e^{n\pi i/6}\sqrt{3})\\ &=2^n3^{(1-n)/2}\Re(e^{n\pi i/6-\pi i/2})\\ &=2^n3^{(1-n)/2}\cos\bigg(\frac{n\pi}{6}-\frac{\pi}{2}\bigg)\\ &=2^n3^{(1-n)/2}\sin\frac{n\pi}{6}\\ \end{align*}