Summing up binomial coefficients

  • #1
utkarshakash
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Homework Statement


The value of [itex]((^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
[/itex]


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
 

Answers and Replies

  • #2
Simon Bridge
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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?
 
  • #3
haruspex
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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.
 
  • #4
utkarshakash
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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
'[itex]\dfrac{(\omega ^2n + \omega ^n)^2}{4} [/itex]

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

[itex](1+\omega)^n + (1+ \omega ^2)^n = 2(C_0+C_3+.........) - (C_1+C_2+C_4...........)[/itex]
 
  • #5
haruspex
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I could simplify the first series to
'[itex]\dfrac{(\omega ^2n + \omega ^n)^2}{4} [/itex]

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

[itex](1+\omega)^n + (1+ \omega ^2)^n = 2(C_0+C_3+.........) - (C_1+C_2+C_4...........)[/itex]
Consider other linear combinations of (1+1)n, (1+ω)n, (1+ω2)n.
 

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