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


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

green line

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 (50th) term is 298. Therefore,

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

green line

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.

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