1. ## Sequence model

Here's one that might/might not be a bit of a challenge.

Construct a model of the periodic series
3, 6, 3, -3, -6, -3, 3, 6, 3, -3, -6, -3, 3, ...

I can construct a model for this, but the student that asked me the question also said that he was doing a unit on geometric and arithmetic series. My model doesn't even remotely resemble this approach. I can't see any way to apply two arithmetic series to this and get a periodic cycle, but is there a way to do this with two geometric series? I can't think of a way to approach this problem generally.

To be specific, could this somehow be something like
$c_n = a_0r^n + b_0p^n$
or some such?

Thanks!
-Dan

(If you are curious I can post my solution. Just ask.)

2. I've been perusing the OEIS files and found that the series 1, 2, 1, -1, -2, 1, ... (my series divided by 3) can be generated by
$c_n = \frac{1}{n!} \left [ \left ( \frac{d}{dx} \right ) ^n \left ( \frac{1 + x}{1 - x + x^2} \right ) \right ]$
evaluated at x = 0. (aka the coefficient of the nth term of the Taylor expansion of (1 + x)(1 - x + x^2) about x = 0.)

I can't seem to develop a pattern for the derivatives. Can anyone think of a way to use this information such that a pre-Calc student might be able to understand it?

Thanks again.
-Dan

3. Well, if I had to guess no one can do this in a "simple" format. Just for the record here is the solution I came up with:

I modeled the sequence 3, 6, 3, -3, -6, -3 with a 5th degree polynomial:
$c_x = -\frac{1}{20} \cdot x^5 + \frac{7}{8} \cdot x^4 - 5x^3 + \frac{77}{8} \cdot x^2 - \frac{49}{20} \cdot x$

Then, to make it a cycle I let $x \equiv n~\text{(mod 6)}$. (I then added a 1 to this in base 10 to allow for the fact that my sequence starts with n = 1, and not n = 0.) As a formula, I get
$x = 6 \left ( \frac{n}{6} - \left \lfloor \frac{n}{6} \right \rfloor \right ) + 1$

So in all its glory:
$c_n = -\frac{1}{20} \cdot \left [ 6 \left ( \frac{n}{6} - \left \lfloor \frac{n}{6} \right \rfloor \right ) + 1 \right ] ^5 +$ $\frac{7}{8} \cdot \left [ 6 \left ( \frac{n}{6} - \left \lfloor \frac{n}{6} \right \rfloor \right ) + 1 \right ] ^4 -$ $5 \cdot \left [ 6 \left ( \frac{n}{6} - \left \lfloor \frac{n}{6} \right \rfloor \right ) + 1 \right ] ^3 +$ $\frac{77}{8} \cdot \left [ 6 \left ( \frac{n}{6} - \left \lfloor \frac{n}{6} \right \rfloor \right ) + 1 \right ] ^2 -$ $\frac{49}{20} \cdot \left [ 6 \left ( \frac{n}{6} - \left \lfloor \frac{n}{6} \right \rfloor \right ) + 1 \right ]$

I don't know if a High Schooler would think of this, but it is at least something that a High Schooler could comprehend.

-Dan