closed form for $\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$

Question

closed form for $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$$

I tried to solve it by :

$$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}=\sum_{k=0}^{n/3}\binom{n}{3k}$$

$$\sum_{k=0}^{n/3}\binom{n}{3k}=\sum_{k=0}^{n/3}\frac{1}{2\pi i}\int_{\left |z \right |=1}\frac{(1+z)^n}{z^{3k+1}}dz$$

then using geometric series but i got no result .

what is your suggest to solve ?

are you sure that $n$ will be $3k$ for some $k\in \Bbb Z$? — Guy, Mar 18 '14 at 19:53
Also for what it's worth, this will be equal to $\Re\left((1+\omega)^n\right)$ where $\omega=e^{2i\pi/3}$ — Guy, Mar 18 '14 at 19:56
Related: How do I count the subsets of a set whose number of elements is divisible by 3? 4? — Grigory M, Mar 18 '14 at 19:59
You were in the right track. You can sum up to $\displaystyle{\large\infty}$ and the integrand becomes $\displaystyle{\large{\left(1 + z\right)^{,,n}z^{2} \over z^{3} - 1}}$ which has three poles. In the integral you must have $\displaystyle{\large\left\vert, z,\right\vert > 1}$ instead of $\displaystyle{\large\left\vert, z,\right\vert = 1}$. — Felix Marin, Jul 28 '14 at 23:27

mookid · Accepted Answer · 2014-03-19T01:20:04.150

2

Assume that $3|n$. $$S_1 = \binom{n}{0}+\binom{n}{3}+\cdots +\binom{n}{n}\\ S_2 = \binom{n}{1}+\binom{n}{4}+\cdots +\binom{n}{n-2}\\ S_3 = \binom{n}{2}+\binom{n}{5}+\cdots +\binom{n}{n-1}\\ S_1 + S_2 + S_3 = 2^n\\ S_1 + jS_2 + j^2 S_3 = (1+j)^n = (-1)^{n/3}\\ S_1 + j^2S_2 + j S_3 = (1+j^2)^n = (-1)^{n/3} $$ with $j^3 = 1, j\notin \Bbb R$.

Now sum up everything: $$ 3S_1 + (1+j+j^2)(S_2+S_3) = 2^n + 2(-1)^{n/3}\\ S_1 = \frac 13[2^n + 2(-1)^{n/3}] $$

edited Mar 19 '14 at 01:20

answered Mar 18 '14 at 19:56

mookid

28,236

what is $j$?${}$ – Guy Mar 18 '14 at 19:57
this is a complex root of $X^3 = 1$ – mookid Mar 18 '14 at 19:57
ah. see my comment below the question. I came up with something similar. +1 – Guy Mar 18 '14 at 19:58
wait. where do you get $(1+j)^3=-1$? – Guy Mar 18 '14 at 20:06
$(1+j)^3 = 1 + 3j + 3j^2 + 1 = 1 - 3 +1 = -1$ because the sum of the roots of $X^3-1$ is 0. – mookid Mar 18 '14 at 20:07
okay nice.${}{}$ – Guy Mar 18 '14 at 20:08
I ended the computation. – mookid Mar 19 '14 at 01:20

closed form for $\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$

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