A Model of Progression: Geometric Series

I’m still thinking about the idea of a model of progression — breaking down a complex activity into a series of manageable components.

I tried to put this idea into practice with the formula for the sum of a geometric series. Here is a proof of the formula, lifted from Purplemath.

I’ve tried to teach this proof before, always pretty unsuccessfully. The long division step seemed like hocus pocus and lost most kids, and they either got lost in the sauce or disengaged completely. Maybe a few kids got something out of it, but it also reinforced student ideas that math doesn’t make sense and is something done to them, rather than with them.

I want students to understand this complex idea, and one essential building block is to understand the following property of polynomial multiplication and division:

So the day before I planned to introduce the geometric series formula, we did an instructional routine drawn from Routines for Reasoning called Recognizing Repetition. I gave students these expressions:

First, I asked students to distribute and rewrite each expression and notice what was being repeated each time. They worked individually at first and then shared ideas at tables, and each table reached some informal ideas about every term cancelling except the first and the last. Then, I asked students to think about generalizations they could make to formalize their thinking into a broader rule. This led to a challenging discussion — again, full of informal ideas but without many resources to write a formal generalization. I ended up doing some of the work to formalize our collective thinking, and we finished with a short meta-reflection on the process of recognizing repetition and writing generalizations.

My goal with this task was to separate the polynomial operations from the rest of the mathematical thinking, get students comfortable with the essential ideas of polynomial operations, and set them up for better success with the proof the next day.

Working through the proof felt much more successful than I’ve ever been before, but with one additional challenge. There is a parity issue, where the difference between n-1 and  n creates some challenges in figuring out exactly how many terms each expression represents and what the appropriate exponent should be in the final formula. It’s tricky — that if a series includes all terms from 0 to n-1, there are actually n terms in that series. While doing the polynomial operations bit ahead of time made a big difference, next time I need to figure out how to focus on that parity question to help the proof go even more smoothly.

More broadly, I really enjoyed doing this type of thinking, and want to figure out what else I can do to lead into challenging topics in a progression that is designed for student understanding. Next up, exponential functions and logarithms!

5 thoughts on “A Model of Progression: Geometric Series”

1. howardat58

Either your transcription is in error or Purplemath got it wrong, but I read it

Sum from 0 to n of “a subscript i” is equal to a(1 – x^n)/(1 – x)
More interesting comment later….

2. howardat58

This is a result for the theorem……

1 + x + x^2 + x^3 + … + x^(n-1) + x^n is in recursive function terms f(n+1) = f(n) + x^n
————— f(n) ——————
but I can rearrange it as 1 + x(1 +x(1 + x(1 + x(1 + x)))) …to term 5 for example,
which gives
f(n+1) = 1 + xf(n) … in general ….
So eliminating f(n+1) we get
sum = (x^n – 1)/(x – 1)

The rearrangement is the general polynomial for early efficient computing circa 1967
Too tricky for the students I fear !

3. Win Smith

If you want a less abstract way to understand the formula for geometric series, try:
Let S be the sum of the series.
Consider xS and see how similar it is to S.
Subtract xS from S, cancelling most terms.
Divide both sides by 1-x.

4. Janelle

Je ne sais pas exactement pourquoi , mais ce blog est de chargement trÃ¨s lent pour moi. Quleuq’un d’autre a ce problÃ¨me ou est-ce un problÃ¨me de mon cÃ´tÃ©? Je vais vÃ©rifier Ã  nouveau plus tard et voir si le problÃ¨me persiste . Cordialement!