# Formula for an arithmetic series (Part 4)

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 an arithmetic series. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

To get the idea across, consider the arithmetic series

$S = 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 + 43$

Now write the sum in reverse order. This doesn’t change the value of the sum, and so:

$S = 43 + 40 + 37 +34+ 31 + 28 + 25 + 22 + 19 + 16$

Now add these two lines vertically. Notice that $16 + 43 = 59$, $19 + 40 = 59$, and in fact each pair of numbers adds to $59$. So

$2S = 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59$

$2S = 59 \times 10 = 590$

$S = 295$

Naturally, this can be directly confirmed with a calculator by just adding the 10 numbers.

When I show this to my students, they often complain that there’s no way on earth that they would have thought of that for themselves. They wouldn’t have thought to set the sum equal to $S$, and they certainly would not have thought to reverse the terms in the sum. To comfort them, I tell them my usual tongue-in-cheek story that this idea comes from 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.

The derivation of the general formula proceeds using the same idea. If $a_1, \dots, a_n$ are the first $n$ terms of an arithmetic sequence, let

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

Recalling the formula for an arithmetic sequence, we know that

$a_2 = a_1 + d$

$\vdots$

$a_{n-1} = a_1 + (n-2)d$

$a_n = a_1 + (n-1)d$

Substituting, we find

$S = a_1 + [a_1 + d] + \dots + [a_1 + (n-2)d] + [a_1 + (n-1)d]$

As above, we now return the order…

$S = [a_1 + (n-1)d] + [a_1 + (n-2)d] + \dots + [a_1 + d] + a_1$

… and add the two equations:

$2S = [2a_1 + (n-1)d] + [2a_1 + d+(n-2)d] + \dots + [2a_1 +(n-2)d+ d] + [2a_1+(n-1)d]$

$2S = [2a_1 + (n-1)d] + [2a_1 + (n-1)d] + \dots + [2a_1 +(n-1)d] + [2a_1+(n-1)d]$

$2S = n[2a_1 + (n-1)d]$

$S = \displaystyle \frac{n}{2} [2a_1 + (n-1)d]$

We also note that the formula may be rewritten as

$S = \displaystyle \frac{n}{2} [a_1 + \{a_1 + (n-1)d\} ]$

or

$S = \displaystyle \frac{n}{2} [a_1 + a_n]$

This latter form isn’t too difficult to state as a sentence: the sum of a series with $n$  is the average of the first and last terms, multiplied by the number of terms.

Indeed, I have seen textbooks offer proofs of this formula by using the same logic that young Gauss used to find the sum $1 + 2 + \dots + 99 + 100$. The “proof” goes like this: Take the terms in pairs. The first term plus the last term is $a_1 + a_n$. The second term plus the second-to-last term is $a_2 + a_{n-1} = a_1 + d + a_n - d = a_1 + a_n$. And so on. So each pair adds to $a_1 + a_n$. Since there are $n$ terms, there are $n/2$ pairs, and so we derive the above formula for $S$.

You’ll notice I put “proof” in quotation marks. There’s a slight catch with the above logic: it only works if $n$ is an even number. If $n$ is odd, the result is still correct, but the logic to get the result is slightly different. That’s why I don’t particularly recommend using the above paragraph to prove this formula for students, even though it fits nicely with the almost unforgettable Gauss story.

That said, for talented students looking for a challenge, I would recommend showing this idea, then point out the flaw in the argument, and then ask the students to come up with an alternate proof for handling odd values of $n$.