Formula for an arithmetic series (Part 7)

As we’ve discussed, the formula for an arithmetic series is

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

where $n$ is the number of terms, $a_1$ is the first term, $d$ is the common difference, and $a_n$ is the last term. This formula may be more formally expressed as

$S = \displaystyle \sum_{k=1}^n a_k = \displaystyle \frac{n}{2} (2a_1 + [n-1] d) = \displaystyle \frac{n}{2} (a_1 + a_n)$

For homework and on tests, students are asked to directly plug into this formula and to apply this problem with word problems, like finding the total number of seats in an auditorium with 50 rows, where there are 12 seats in the front row and each row has two more seats than the row in front of it.

In my opinion, the ability to solve questions like the one below is the acid test for determining whether a student — who I assume can solve routine word problems like the one above — really understands series or is just familiar with series. In other words, if a student can solve routine word problems but is unable to handle a problem like the one below, then there’s still room for that student’s knowledge of series to deepen.

Calculate $\displaystyle \sum_{k=11}^{60} (5k - 2)$

There are two reasonable approaches for solving this problem.

Solution #1. Notice that $5k - 2 = 5(k-1) + 5 - 2 = 3 + 5(k-1)$ . So this is really an arithmetic series whose first term is $3$ and whose common difference is $5$ . Therefore,

$S = \displaystyle \sum_{k=1}^{60} a_k = \displaystyle \frac{60}{2} (2[3] + [60-1] 5)=9030$

However, I’m supposed to start the series on $k=11$ , not $k=1$ . That means that I need to subtract off the first ten terms of the above series. Now

$S = \displaystyle \sum_{k=1}^{10} a_k = \displaystyle \frac{10}{2} (2[3] + [10-1] 5)= 255$

Finally,

$\displaystyle \sum_{k=11}^{60} a_k = \displaystyle \sum_{k=1}^{60} a_k - \displaystyle \sum_{k=1}^{10} a_k = 9030 - 255 = 8775$

Solution #2. Writing out the terms, we see that

$\displaystyle \sum_{k=11}^{60} (5k - 2) = (5[11]-2) + (5[12]-2) + \dots + (5[60]-2)$

$\displaystyle \sum_{k=11}^{60} (5k - 2) = 53+58 + \dots +298$

The right-hand side is an arithmetic series whose “first” term is $53$ and whose last ( $50$ th) term is $298$ . Therefore,

$\displaystyle \sum_{k=11}^{60} (5k - 2) = \frac{50}{2} (53+298) = 8775$

Of the two solutions, I suppose I have a mild preference for the first, as the second solution won’t work for something like $\displaystyle \sum_{k=11}^{60} k^2$ . However, both solution demonstrate that the student is actually thinking about the meaning of the series instead of just plugging numbers in a formula, and so I’d be happy with either one in a Precalculus class.

Formula for an arithmetic series (Part 6)

In the previous posts of this series, I described two methods of deriving the formula

$\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$

The first method concerned reversing the terms of the sum (or, almost equivalently, taking the terms in pairs). The second method used mathematical induction.

Mathematical induction can be applied to arithmetic series as well as other series. However, the catch is that you have to know the answer before proving that the answer actually is correct. By contrast, the first method did not require us to know the answer in advance — it just fell out of the calculation — but it cannot be applied to series that are not arithmetic.

Here’s a third method using the principle of telescoping series. This method has the strengths of the previous two methods: it does not require us to know the answer in advance, and it can also be applied to some other series which are not arithmetic.

To begin, consider the sum

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2]$

At this early point, students often object, “Where did that come from?” I’ve said it before but I’ll say it again: 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.

In any event, I will evaluate this sum in two different ways.

Step 1. Just write out the terms of the series, starting from $k=1$ and ending with $k =n$ .

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = [1^2 - 0^2] + [2^2 - 1^2] + [3^2 - 2^2] + \dots + [n^2 - (n-1)^2]$

Notice that, on the right-hand side, the $1^2$ terms cancel, the $2^2$ terms cancel, and so on. In fact, almost everything cancels. The only two terms that aren’t cancelled are the $0^2$ and $n^2$ terms. Therefore,

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = n^2 - 0^2 = n^2$

Step 2. Next, we’ll rewrite the original sum by expanding out the terms inside of the sum:

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n [k^2 - (k^2 -2k + 1)]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n [2k-1]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n 2k - \displaystyle \sum_{k=1}^n 1$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 2\displaystyle \sum_{k=1}^n k - \displaystyle \sum_{k=1}^n 1$

Step 3. Of course, these different looking answers from Steps 1 and 2 have to be the same, so let’s set them equal to each other:

$2\displaystyle \sum_{k=1}^n k - \displaystyle \sum_{k=1}^n 1 = n^2$

There is one unknown in this equation, $\displaystyle \sum_{k=1}^n k$ . The second sum is just the constant $1$ added to itself $n$ times, and so $\displaystyle \sum_{k=1}^n 1 = n$ . Therefore, we solve for the unknown: