Formula for a finite geometric series (Part 8)

As I’ve said before, I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

While I’m not a fan of making students memorize formulas, I am a fan of teaching students how to derive formulas. Speaking for myself, if I ever need to use a formula that I know exists but have long since forgotten, the ability to derive the formula allows me to get it again.

Which leads me to today’s post: the derivation of the formulas for the sum of a finite geometric series. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

green lineLike its counterpart for arithmetic series, the formula for a finite geometric series can be derived using the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

If a_1, \dots, a_n are the first n terms of an geometric sequence, let

S = a_1 + a_2 + a_3 + \dots + a_{n-1} + a_n

Recalling the formula for an geometric sequence, we know that

a_2 = a_1 r

a_3 = a_1 r^2

\vdots

a_{n-1} = a_1 r^{n-2}

a_n = a_1 r^{n-1}

Substituting, we find

S = a_1 + a_1 r+ a_1 r^2 \dots + a_1 r^{n-2} + a_1 r^{n-1}

At this point, we use something different from the patented Bag of Tricks: we multiply both sides by -r.

-rS = -a_1r - a_1 r^2- a_1 r^3 \dots - a_1 r^{n-1} - a_1 r^n

Next, we add the two equations. Notice that almost everything cancels on the right-hand side. The a_1 r cancel, the a_1 r^2 cancel, yada yada yada, and the a_1 r^{n-1} cancel. The only terms that remain are a_1 and -a_1 r^n. So

S - rS = a_1 - a_1 r^n

S(1-r) = a_1 (1- r^n)

S = \displaystyle \frac{a_1 ( 1-r^n) }{1-r}

A quick pedagogical note: I find that this derivation “sells” best to students when I multiply by -r and add, as opposed to multiplying by r and subtracting.

green line

This formula is also a straightforward consequence of the factorization formula

x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + \dots +x y^{n-2} + y^{n-1})

Just let x=1 and y=r, and then multiply both sides by the first term a_1.

However, in my experience, most students don’t have instant recall of this formula either. They can certainly remember the formula for the difference of two squares (which is a special case of the above formula), but they often can’t remember that the difference of two cubes has a formula. (And, while I’m on the topic, they also can’t remember that the sum of two cubes can always be factored.)

green line

Pedagogically, the most common mistake that I see students make when using this formula is using the wrong exponent on the right-hand side. For example, suppose the problem is to simplify

48 +24+ 12 + 6 + 3 + 1.5 + 0.75 + 0.375

Here’s the common mistake: student solve for n using the formula for a geometric sequence. They solve for the unknown exponent (often using logarithms) and find that 0.375 = 48 (0.5)^7. They conclude that n=7, and then plug into for formula for a geometric series:

S = \displaystyle \frac{48 ( 1-(0.5)^7) }{1-0.5} = 95.25 (incorrect)

This answer is clearly wrong, since the sum of the original series must have a 5 in the thousandths place. The answer 95.25 is the correct answer to the wrong question — that’s the sum of the first seven terms of the sequence (stopping at 0.75), but the original series has eight terms. Using the formula correctly, we find

S = \displaystyle \frac{48 ( 1-(0.5)^8) }{1-0.5} = 95.625 (correct)

Not surprisingly, the difference between the incorrect and correct answers is 0.375, the eighth term.

To help students avoid this mistake, I re-emphasize that the number n stands for the number of terms in the series. In particular, it does not mean the exponent needed to give the last term in the series. That exponent, of course, is n-1, not n.

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