# Opening Question

I had a wonderful DO NOW in my Advanced Precalculus Class that allowed me to have an insight that I haven’t had before.

Sum the following:

(a) 3+6+9+12+…+300

(b) 4+7+10+13+…+301

For some background, during our last unit, students worked on The Fistbump Problem, and they eventually figured out how to add 1+2+3+…+n.

I wasn’t sure if students would make the connection to that problem and the first sum. A few groups did, and rewrote it as:

3(1+2+3+…+100)

And then they did the sum the knew how to (from the fistbump problem), and tripled it.

Others did the “pairing” method where they said 3+300=303, 6+297=303, 9+294=303, … but got a little stuck on figuring out how many pairs they had.

Regardless, they had the appropriate insights.

I love question (b) because even if students didn’t get (a), they should have been able to say the answer to (b) in relation to the answer to (a). So I threw up the answer to part (a) that students had provided, and said: “Okay, let’s say you wanted to quickly get the answer to (b) without doing an arduous process… What do we do?”

Groups had 30 seconds to think and talk.

Almost all saw the answer to (b) was 100 more than the answer to (a) — because you’re secretly adding 1 to each of the terms in (a) to get the terms in (b), and there are 100 terms!

***

These two questions gave me a whole new way to think about all arithmetic series… as “transformations” of the series 1+2+3+…+n.

For example:

If I want to add 5+8+11+14+17+20+23+26+29+32+35, I could do the following

First find out how many terms there are in the sum (in this case, 11)

Sum the series 1+2+3+4+5+6+7+8+9+10+11=66 (this is the one thing kids need to know how to do)

Triple the sum to get the correct “common difference”): 3+6+9+12+…+33=66(3)

Add 2 to each term to get the sum I want: 5+8+11+14+…+35=66(3)+2(11)

Done!

I have to think some more about this, but I really think I’m onto a new way for students to look at arithmetic sequences as a transformation of the most basic arithmetic sequence. The transformation is a stretch (multiplication by common difference) and a shift (addition of a fixed value to each term). Whether or not the new way gives students anything better or more useful than the approach I currently use with them is a different story. But I love that *I* had this new insight about something I’ve been teaching for years.

UPDATE: I forgot to mention that I showed it to another math teacher, who said, well, yes, it’s just:

$\sum (3x+2) = 3\sum (x)+\sum(2)$

which sort of took the neatness out of it — probably because of the math-y symbols. But then I realized that the thrill I got out of it was the conceptual insight, not the algebraic insight. And I figure students who might be able to discover this on their own will get the same thrill. If you spend a lot of time with the summation sign, what this approach might allow is have students discover the rules of using the summation sign.

I love one mathematical motto of class being take what you don’t know and turn it into what you do know. That undergirds lots of what we do in class. And that is precisely what this approach is doing!

## 8 thoughts on “Opening Question”

1. Now all that’s left is to do the algebra from the verbal description.

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2. The sum of an arithmetic sequence with n terms can also be thought of as: n multiplied by the average entry in the sequence. (I will not “explain” why, but rather leave that Fact to dangle!)

For example, if n = 100, and you have: 1 + 2 + … + 100:
The average is 50.5. And so the total is 100×50.5 = 5050.

Multiplying the entire sequence by a constant c leads to a new sequence whose average entry is: c multiplied by the previous average.

Similarly, adding a constant k to each of the terms shifts the average in the new sequence by k.

This could provide another inroad for investigating the phenomenon that you discovered!

For related remarks, see my earlier betterQs post:
Take-Homes from Gauss & Questions About Questions

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• Maya, now you’ve done it, minus the symbols!

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Yes! Your observations are precisely the connections that the statistics teacher made when I showed him this! Thank you for noting it here — because surely by the time next year comes about, I will have forgotten this additional vantage point!

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• You are welcome. (And I agree that a conceptual insight is more exciting than a notational one!)

One more thought: Your observations around a “stretch” and a “shift” are essentially what a linear transformation does. (In fact, you use the word ‘transformation’.) For example, the function f(x)=mx+b “stretches” by a factor of m and “shifts” by b.

Sometimes, arithmetic sequences are used to scaffold towards linear functions. For example, a sequence like: 3, 5, 7, … might be encountered as a precursor (of sorts) for the function defined by f(x)=2x+1. And so one way to think about moving between various arithmetic sequences is by transforming 1, 2, …, n with a “stretch” (to account for the arithmetic difference between terms) and a “shift” (to account for the first term in the sequence).

Succinctly: You might think of the “1, 2, …, n” arithmetic sequence as analogous with “f(x)=x”; in this way, the different arithmetic sequences become analogous with the different linear functions. Depending on your perspective, such an analogy might re-enforce the importance of conceptually grasping “1, 2, …, n” or “f(x)=x” (or both!).

[I apologize if this is all old hat — you remark on the possibility of losing a vantage point by next year, and so I thought this might be worth recording.]

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4. suevanhattum

Maya’s last point is why I teach arithmetic series with linear functions and geometric series with exponential functions. But Sam, I love your insight. Every time I have a big insight like that, it makes me love teaching the topic.

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5. Incidentally, there’s a lovely sequence of questions that develops things this way at the beginning of the Art of Problem Solving’s “Introduction to Combinatorics” book, but instead of adding we’re just trying to count how many terms are in the sequence. For example, to count how many numbers appear in the list 3, 6, 9, 12, …, 366, the first thing you do is divide by 3, turning it into 1, 2, 3, …, 122. Now it’s clear there are 122 items in the list.

Similarly (and in some ways, more challenging), if you have the list 6, 7, 8, 9, 10, …, 191, most students assume there are 185 items in the list just by subtracting 191-6. But if you actually subtract 5 from each term, you see that the list is equivalent to 1, 2, 3, …, 186. So there are actually 186 numbers there!

In general, it seems like whenever you have a big list of numbers (sequence or series), one methodical way to do things is to transform it to a list starting at 1, increasing by 1 each time. I thought that was pretty cool.

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