Summing up binomial coefficients

[utkarshakash]

Homework Statement

The value of ((^n C_0+^nC_3+...) - \frac{1}{2} (^nC_1+^nC_2+^nC_4+^nC_5+...))^2 + \frac{3}{4} (^nC_1-^nC_2+^nC_4-^nC_5...)^2<br />

The Attempt at a Solution

I can see that in the left parenthesis, the first bracket contains terms which are multiples of 3 and in the second bracket, those terms are missing. I know it's a much less attempt from my part but that's all I can see. Please help XO

 

[Simon Bridge]

Sums go to n?
Have you tried expanding out the squares?
You'll need extra binomial coefficients.

The coefficients have a symmetry amongst other properties - you should make a cheat-sheet of the basic properties.

Also - try n=2 and n=3 and expand them out explicitly and see if you find a pattern.

BTW: what was the question?

 

[haruspex]

Try using powers of a cube root of 1. E.g. look at how the imaginary parts of those change and compare it with the behaviour of the last series.

 

[utkarshakash]

Try using powers of a cube root of 1. E.g. look at how the imaginary parts of those change and compare it with the behaviour of the last series.

I could simplify the first series to
'\dfrac{(\omega ^2n + \omega ^n)^2}{4}

But I still can't get the second one. Here's my attempt

(1+\omega)^n + (1+ \omega ^2)^n = 2(C_0+C_3+...) - (C_1+C_2+C_4...)

 

[haruspex]

I could simplify the first series to
'\dfrac{(\omega ^2n + \omega ^n)^2}{4}

But I still can't get the second one. Here's my attempt

(1+\omega)^n + (1+ \omega ^2)^n = 2(C_0+C_3+...) - (C_1+C_2+C_4...)

Consider other linear combinations of (1+1)ⁿ, (1+ω)ⁿ, (1+ω²)ⁿ.