View attachment 34760
First two are easy, can someone help me out with the third one - not getting a start...
Is
![](https://latex.codecogs.com/png.latex?\bg_white \left(-1\right)^{n}=\left(1+\omega\right)^{3n}={3n\choose 0}1^{3n}+{3n\choose 1}\omega+{3n\choose 2}\omega^{2}+{3n\choose 3}\omega^{3}+...)
noting that
![](https://latex.codecogs.com/png.latex?\bg_white \omega^{3}=1)
and that
![](https://latex.codecogs.com/png.latex?\bg_white \omega^{3n}=1)
for
Now diagramatically we will use the fact that
![](https://latex.codecogs.com/png.latex?\bg_white \omega^{2})
is the complex conjugate of
![](https://latex.codecogs.com/png.latex?\bg_white \omega)
and as such
![](https://latex.codecogs.com/png.latex?\bg_white \omega^{2}=\bar{\omega})
.
With this is hand we know that a complex number plus its conjugate gives us two times the real part. Unfortunately we need to satisfy the fact that
![](https://latex.codecogs.com/png.latex?\bg_white \omega+\omega^{2}=-1)
so therefore what you see here is the involvement of the
![](https://latex.codecogs.com/png.latex?\bg_white -\frac{1}{2})
so as such for the
![](https://latex.codecogs.com/png.latex?\bg_white {3n\choose 1} + {3n\choose 2} + {3n\choose 4} + {3n\choose 5})
pattern there needs to be a
![](https://latex.codecogs.com/png.latex?\bg_white -\frac{1}{2})
to satisfy what we know from part i. Thus, after the dust is cleared
![](https://latex.codecogs.com/png.latex?\bg_white {3n\choose 0}-\frac{1}{2}\left({3n\choose 1}+{3n\choose 2}\right)+{3n\choose 3}-\frac{1}{2}\left({3n\choose 4}+{3n\choose 5}\right)+{3n\choose 6}+{3n\choose 3n}=\left(-1\right)^{n})