Simplify without using complex numbers.

Question

Simplify $$\large\sum_{r=0}^{\left \lfloor \frac{n}{3} \right \rfloor}\binom {n}{3r}$$

I tried as much as I could; tried to apply induction, tried to approach combinatorally but failed. I could not resist to see the solution after trying it for all the day. But the answer used complex numbers(cube root of unity and de’moiver’s) and I didn’t find it elegent anyway. In short, I am asking you to simplify it without using complex numbers.

I have asked this on AOPS forum too but people say it cannot be solved without using complex numbers. Please help me!

It can be simplified using a purely combinatorial proof. See the proof of Theorem 1.1 in Third and Fourth Binomial Coefficients, from the Fibonacci Quarterly, Volume 49, Number 2. — Mike Earnest, Oct 29 '19 at 17:04
I find the complex numbers proof really elegant, though. What bothers you about it? — ViHdzP, Oct 29 '19 at 17:23
Possible duplicate of How do I count the subsets of a set whose number of elements is divisible by 3? 4? — Grigory M, Nov 03 '19 at 16:10
Without more context for what kinds of proof you would accept (not liking complex numbers doesn't give Readers much appreciation what you do like), I'd have to vote to close for lack of context. — hardmath, Nov 04 '19 at 01:04

Samdney · Answer 1 · 2019-10-29T19:43:21.887

0

If you assume the case $n:=3m$, $m \in \mathbb{N}_{0}$, you can write

$\sum_{r = 0}^m {3m \choose 3r}$

Wolframalpha gives for this:

$\frac{1}{3}\left(2(-1)^{m} + 8^{m}\right)$

Wolframalpha Calculation

I'm not completly able to reproduce this result, but this is maybe a good starting point.

edited Oct 29 '19 at 19:43

answered Oct 29 '19 at 19:07

Samdney

61

Simplify without using complex numbers.

1 Answers1