Find the sum of the series

$$\sum_{r=0} ^n \binom{6n}{3r} $$

My attempt... I tried generating a binomial half series but since the question is actually the sum of a kind of one-sixth series....I have no idea on how to proceed...I also tried using calculus but it got me nowhere.....Can anyone help me out? Thanks in advance!!


You know that $f(x)=(x+1)^n$ has expansion $\sum_{i=0}^n \binom{n}{i}x^i$. Let $\omega$ be a cubic primitive root of unit. Then $$ \sum_{i=0}^{\lfloor n/3\rfloor} \binom{n}{3i}=\frac{f(\omega)+f(\omega^2)+f(\omega^3)}{3}=\frac{(1+\omega)^n+(1+\omega^2)^n+2^n}{3}. $$

In your example, setting $6n$ and considering that $1+\omega+\omega^2=0$, you get $$ \frac{(1+\omega)^{6n}+(1+\omega^2)^{6n}+2^{6n}}{3}=\frac{(-\omega^2)^{6n}+(-\omega)^{6n}+2^{6n}}{3}=\frac{2^{6n}+2}{3}. $$

Ps. Of course, you can obtain the same result by setting mechanical resursions.

Ps2. As correctly noted by Marko below, the original summation goes up to $n$, not $2n$. Hence, exploiting the symmetry of the binomials, we have $$ \sum_{i=0}^{n}\binom{6n}{3i}=\frac{1}{2}\left(\sum_{i=0}^{2n}\binom{6n}{3i}\,+\binom{6n}{3n}\right)=\frac{2^{6n-1}+1}{3}+\frac{1}{2}\binom{6n}{3n}. $$

  • $\begingroup$ Did you verify your formula on some sample values? I get two different sequences namely $21, 1145, 68001, 4148281, 256515731, 15990813773,\ldots$ and $22, 1366, 87382, 5592406, 357913942, 22906492246,\ldots$ $\endgroup$ – Marko Riedel Oct 11 '16 at 20:39
  • $\begingroup$ @MarkoRiedel I didn't try sample values, but at least for $n=1$ it seems to be correct, isn't it? $\endgroup$ – Paolo Leonetti Oct 11 '16 at 20:41
  • $\begingroup$ Pluggin in $6n$ in your formula you get $2n$ for the upper limit and not $n.$ $\endgroup$ – Marko Riedel Oct 11 '16 at 20:42
  • $\begingroup$ @MarkoRiedel I added the truncated sum, it seems to work now ;) $\endgroup$ – Paolo Leonetti Oct 11 '16 at 20:53
  • $\begingroup$ Glad to see you again, and (+1) for the beautiful answer, of course. $\endgroup$ – Jack D'Aurizio Oct 11 '16 at 22:10

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