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My teacher was discussing binomial expansions of $(1 + x)^n$ and he gave as an interesting example with $x = i$ whereby you could obtain the sum of all the odd coefficients ($C_n^1+ C_n^3+ C_n^5 ...$) and the even ones. Then by appying deMoivre to $(1 + i)^n$ you could separate the binomial expansion into a real and an imaginary part and compute those separately.

How can I, in a somewhat similar manner, determine the sum of every kth binomial coefficient, using the kth unity root?

By substituting into $(1 + x)^n$ all of the unity roots on by one and then adding the results the results is the sum of every k-th coefficient is equal to $(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n$ but how can I get the closed form for this sum?

andreas.vitikan
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1 Answers1

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The keyword is discrete Fourier transform. The basic lemma is that if $\zeta_k$ is a primitive $k^{th}$ root of unity then

$$\sum_{i=0}^{k-1} \zeta_k^{ij} = \begin{cases} 0 \text{ if } k \nmid j \\ k \text{ otherwise}. \end{cases}$$

It follows that if $f(x) = \sum a_i x^i$ is a formal power series then the sum of every $k^{th}$ term of $f$ is

$$\sum a_{ki} x^{ki} = \frac{1}{k} \sum_{i=0}^{k-1} f(\zeta_k^i x).$$

In this particular case we have, for example,

$$\sum {n \choose 2i} = \frac{1}{2} \left( (1 + 1)^n + (1 - 1)^n \right) = 2^{n-1}$$

and

$$\sum {n \choose 3i} = \frac{1}{3} \left( (1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n \right) = \frac{2^n + (-\omega^2)^n + (-\omega)^n}{3}$$

where $\omega$ is a primitive third root of unity (and we used the fact that $1 + \omega + \omega^2 = 0$). The answer is messier in general.

Qiaochu Yuan
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  • I'm in a high school class, we haven't even done integrals and derivatives. There must be a somewhat easier way without requiring FTs. – andreas.vitikan Mar 14 '13 at 05:36
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    @andreas: you don't need to know anything about integrals and derivatives to understand the DFT. It's just a name. The technique you want is the DFT whether or not you want to call it that, and to understand it all you need to do is play around with complex numbers a little. – Qiaochu Yuan Mar 15 '13 at 07:28
  • More generally, $\sum_{n\in\mathbb{N}}a_nx^n=f(x)$ implies $\sum_{n\in m\mathbb{N}}a_nx^{nl+o}=\frac{x^o}{m}\sum_{i\in[m]}f(\omega^ix^l)$. – Leo Apr 26 '15 at 01:02
  • @QiaochuYuan, what about 4$i$ – Mclalalala Mar 04 '17 at 08:52