Prove ${n \choose 1}- \frac{1}{3}{n \choose 3}+ \frac{1}{9}{n \choose 5}-\frac{1}{27}{n \choose7}+...= \frac{2^n}{3^{\frac{n-1}{2}}}\sin({\frac{n\pi}{6}})$
I used the binomial expansion $(1+x)^n$ where I put x = $i$ , so
$(1+i)^n={n \choose 0}+{n \choose 1}i+ {n \choose 2}i^2+{n \choose3}i^3+...$ (i)
I know $(1+i)^n=2^{\frac{n}{2}}(\text{cis}(n\cdot45^{o}))$
Then , I used the expansion $(1-i)^n$ (ii)
(ii) - (i)
${n \choose 1}- {n \choose 3}+ {n \choose 5}-{n \choose7}+... = 2^{\frac{n}{2}}\sin({\frac{n\pi}{4})}$
How can I relate the coefficients $\frac{1}{3^k}$
Hint:$$\left(1+\frac{i}{\sqrt{3}}\right)^n=\left(\binom{n}{0}-\binom{n}{2}+..\right)+\frac{i}{\sqrt{3}}\left(\binom{n}{1}-\binom{n}{3}\frac{1}{3}+...\right)$$