Mathematical Memorandum: Binomial coefficient and imaginary number

In the last part time job, I find the problem that we compute the following sum of binominal coefficients.

$\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots + (-1)^n \binom{n}{n}$

As you know, we can compute by using the binominal theorem. The answer is 0. The last result implies the following formula:

$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots + \binom{n}{2j} = 2^{n - 1}.$

Here, 2j is the greatest number which is not more than n. Then, we can also compute:

$S_{n, k} := \binom{n}{0} + \binom{n}{k} + \binom{n}{2k} + \cdots + \binom{n}{jk} \textrm{ for } k \geq 2.$

Here, jk is the greatest integer which is not more than n. The computation of this is a littel tricky. But we can compute by the following binominal expansions.

$(1 + 1)^n, (1 + \omega )^n, \cdots, (1 + \omega^{k - 1} )^n$

Here,

$\omega := \exp\left ( \frac{2 \pi }{k} i \right ).$

Indeed, since

$1 + \omega^l + \omega^{2l} + \cdots + \omega^{(k - 1)l} = 0 \textrm{ for every } l \geq 1 \textrm{ such as } k \nmid l ,$

summing each expansion, we have the result:

$S_{n, k} = \frac{(1 + 1)^n + (1 + \omega )^n + \cdots + (1 + \omega^{k - 1} )^n}{k}.$

Binomial coefficient and imaginary number

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