1
$\begingroup$

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!!

$\endgroup$
2
$\begingroup$

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 $$ \sum_{i=0}^{n}\binom{6n}{3i}=\frac{1}{2}\left(\sum_{i=0}^{2n}\binom{6n}{3i}\,+\binom{6n}{3n}\right)=\frac{2^{6n-1}+1}{3}+\frac{1}{2}\binom{6n}{3n}. $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.