Going over last night’s homework, make sure they got the right formulas. For an arithmetic series,

### Example 1

Students sometimes ask if that formula will still work for an arithmetic series with an odd number of terms. Obviously, you can’t still pair them up in the way we have been doing. The answer is, it does still work. One proof—which I usually don’t mention unless the right questions are asked—is that, for an arithmetic series, the average of all the terms is right in the middle of the first and last terms. (It can take a minute to convince yourself that this is not always true for any series, but it is for any *arithmetic* series.) So the average is

Anyway, on to today’s topic.

### Note:

*very slowly and carefully*!

Today, we’re going to learn a new way of proving things. This method, called “proof by induction,” is a very powerful and general technique that turns up in many different areas of mathematics: we are going to be applying it to series, but the real point is to learn the technique itself.

So...to begin with, we are going to prove something we already know to be true:

Of course we know how to prove that using the arithmetic series trick, but we’re going to prove it a *different* way.

Let’s start by seeing if that formula works when

How about the right side? Well, that’s...

To build up to the next step, ask this hypothetical question. Suppose we had not yet proven that this equation *always* works. But suppose that I had proven that it works *when*

Well, you can do the right side easily enough:
*your* calculator? No, you don’t, if you’re clever. (See if they can figure this next part out—this is the key.) I already told you what the first 200 numbers add up to. So you can simply add 201 to my total.

The point here is not just “it works.” The point is that you can confirm that it works, *without adding up all 200 numbers again*, because I already did that part—all you have to add is the last number.

Now...suppose I had already proven that it works for *should* get for

Now...suppose I had already proven that it works for *should* get for *what is the general form of this question*? See if they can figure out that it is:

Suppose I had already proven that it works for some

Give them time here...see if they can find the answer...

We would add *should* get for

What does that look like? Well, for the old *should* we get? Well, for

Do the algebra to show that they are equal. Then, step back and say...so, what have we done? Well, first we proved that the formula works for *if it works for any number, it must also work for the next number*. If it works for

At this point, I think it’s helpful to work through one more example. I recommend going through

**Homework:**

“Homework—Proof by Induction”

At this point, you’re ready for the test. Unlike most of my “Sample Tests,” this one is probably too *short*, but it serves to illustrate the sorts of problems you will want to ask, and to remind the students of what we’ve covered.

Comments:"This is the "teacher's guide" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing […]"