Find the value of $\sum\binom{6n}{2k-1}(-3)^k$ 
Find the value of $\sum_{k=1}^{3n}\binom{6n}{2k-1}(-3)^k$.  

My working: 
\begin{align}
&  \sum\binom{6n}{2k-1}(-3)^k\\
=& \sum\binom{6n}{2k-1}(i\sqrt3)^{2k}\\
=& i\sqrt3\sum\binom{6n}{2k-1}(i\sqrt3)^{2k-1}\\
=& \frac{i\sqrt3[(1+i\sqrt3)^{6n}-(1-i\sqrt3)^{6n}]}{2}\\
=& \frac{i\sqrt3[(1+i\sqrt3+1-i\sqrt3)^{3n}((1+i\sqrt3-1+i\sqrt3)^{3n}]}{2}\\
=& \frac{i\sqrt3\cdot2^{3n}2^{3n}(i\sqrt3)^{3n}}{2} \\
=& \frac{2^{6n}(i\sqrt3)^{3n+1}}{2}
\end{align}
But the correct answer is $0$ and I can't figure out how one gets that. Moreover, I want to know what's wrong with my solution.
 A: Hint: $$(1+i\sqrt 3)^{6n}-(1-i\sqrt 3)^{6n}=(re^{i\pi/3})^{6n}-(re^{-i\pi/3})^{6n}=r^{6n}(e^{i2\pi n}-e^{-i2\pi n})=r^{6n}(1-1)=r^{6n}\cdot 0$$
A: $\sum \binom{m}{2k-1}x^{2k-1}
+\sum \binom{m}{2k}x^{2k}
=\sum \binom{m}{k}x^k
=(1+x)^m
$.
$(1-x)^m
=\sum \binom{m}{k}(-1)^kx^k
$
so
$(1+x)^m+(1-x)^m
=\sum \binom{m}{k}x^k(1+(-1)^k)
=2\sum \binom{m}{2k}x^{2k}
$
and
$(1+x)^m-(1-x)^m
=\sum \binom{m}{k}x^k(1-(-1)^k)
=2\sum \binom{m}{2k-1}x^{2k-1}
$.
Therefore,
putting $x = i\sqrt{3}$
and $m = 6n$,
$\begin{array}\\
\sum \binom{6n}{2k-1}(-3)^{k}
&=\sum \binom{6n}{2k-1}(i\sqrt{3})^{2k}\\
&=i\sqrt{3}\sum \binom{6n}{2k-1}(i\sqrt{3})^{2k-1}\\
&=\frac{i\sqrt{3}}{2}((1-i\sqrt{3})^{6n}-(1+i\sqrt{3})^{6n})\\
&=0\\
\end{array}
$
since
$(1-i\sqrt{3})^3
=(1+i\sqrt{3})^3
=-8
$
and
$(1-i\sqrt{3})^6
=(1+i\sqrt{3})^6
=64
$.
Similarly,
$\begin{array}\\
\sum \binom{6n}{2k}(-3)^{k}
&=\sum \binom{6n}{2k}(i\sqrt{3})^{2k}\\
&=\frac{1}{2}((1-i\sqrt{3})^{6n}+(1+i\sqrt{3})^{6n})\\
&=\frac{1}{2}(64^n+64^n)\\
&=64^n\\
\end{array}
$
