Sum of Arithmetic Series | Free Discrete Mathematics Lesson (4 College Credits) | UPI Study

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Sum of Arithmetic Series

Discrete Mathematics · Chapter 3: Sequences and SeriesSection 3.6 · lesson 22 of 879 min · free with a free account

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In this lesson

A whole sum collapsed into one closed-form expression.

  • Define an arithmetic series
  • Derive its sum by writing it forwards and backwards
  • Apply both forms of the sum formula
  • Compare the closed form with repeated addition

Transcript

Everything said in the video, section by section (1,315 words). Read along, or search it.

Sequences and Series

Sum of Arithmetic Series

Section three point five introduced sigma notation as a compact name for a sum. This section goes one step further, for an especially common kind of sum. When the terms form an arithmetic sequence, the whole sum collapses into a single closed-form expression. No loop, and no term-by-term addition. We'll derive it by writing the sum forwards and backwards, apply both forms of the formula, and compare it with repeated addition.

The Opening Problem

A Fiber Rollout's Grand Total

A regional internet provider is planning a multi-year fiber rollout. Crews activate more new connections each month than the month before, as trained crews are added and earlier street work matures. It's a steady, fixed increase, month over month. Sales forecasting doesn't want the month-by-month figures. It wants the grand total across the whole rollout, recomputed instantly every time a planning assumption changes. Re-adding dozens of numbers by hand, every time an assumption shifts, is exactly the kind of repeated, mechanical work discrete mathematics is good at eliminating.

Definition 3.6.1

Arithmetic Series

Recall that an arithmetic sequence has a fixed common difference d between consecutive terms. The i-th term is a one plus i minus one times d. Adding the terms of a finite arithmetic sequence produces its arithmetic series. S sub n is the sum of its n terms, a one through a n. In the notation of the previous section, it's the n-th, and final, partial sum. Adding the terms one at a time always works. But it costs n minus one additions, however the terms are structured. Because each term is just a one plus a fixed multiple of d, that structure makes a dramatic shortcut possible.

A Well-Known Story

1 + 2 + ... + 100

A well-known story in mathematical folklore, whose accuracy is disputed, holds that a young Carl Friedrich Gauss, was asked to add the whole numbers from one to one hundred. He produced the answer almost immediately. Instead of adding in order, he paired the first number with the last, the second with the second-to-last, and so on. One plus one hundred, two plus ninety-nine, three plus ninety-eight. Every one of the fifty pairs sums to one hundred one. Fifty pairs of one hundred one is five thousand fifty, with no adding in sequence.

Step 1

Write the Sum Forwards and Backwards

Here's the general argument, for any arithmetic series of n terms. Write the sum forwards, a one plus a two, up to a n. Since order doesn't matter in addition, write the same total backwards, a n plus a n minus one, down to a one. Now add the two expressions term by term. First with first, second with second, and so on. Two S sub n becomes a sum of exactly n bracketed pairs. The k-th pair is a sub k plus a sub n plus one minus k.

Step 2

Every Pair Sums to the Same Value

Now prove that every pair sums to the same value. Using the formula for a term, a sub k is a one plus k minus one times d. And a sub n plus one minus k is a one plus n minus k times d. Add them. Combining like terms, the k and minus k cancel, leaving two a one plus n minus one times d. And that's a one, plus a one plus n minus one times d. Which is a one plus a n. The crucial point is that the index k canceled out entirely. The value a one plus a n doesn't depend on which pair is examined. So every one of the n pairs sums to exactly that.

Theorem 3.6.1

The Sum of an Arithmetic Series

So two S sub n is n pairs, each equal to a one plus a n. Divide both sides by two. The theorem. For an arithmetic series of n terms, S sub n equals n over two, times a one plus a n. Substituting a n equals a one plus n minus one times d gives the equivalent form, n over two, times two a one plus n minus one times d. Both forms compute the sum with a fixed number of operations, however large n is. As a check, Gauss's problem has a one equal to one, a n equal to one hundred, and n equal to one hundred. That gives fifty times one hundred one, which is five thousand fifty.

Figure 3.6.1

Pairing From the Outside In

This figure shows an eight-term arithmetic series, starting at two, with common difference three. The terms are two, five, eight, eleven, fourteen, seventeen, twenty, and twenty-three. Pair them from the outside in. Two with twenty-three, five with twenty, eight with seventeen, eleven with fourteen. Every pair sums to twenty-five, which is the first term plus the last. That's exactly the fact proved algebraically, here for one specific sequence.

Worked Example 3.6.1

The Rollout Total

Back to the fiber rollout. Suppose month one activates sixty connections, and the monthly figure grows by twenty-five each month, for twelve months. First find the last term. a twelve is sixty plus eleven times twenty-five, which is three hundred thirty-five connections in month twelve. Then apply the theorem. S twelve is twelve over two, times sixty plus three hundred thirty-five. That's six times three hundred ninety-five, or two thousand three hundred seventy connections over the year. And when an assumption changes, the total is recomputed in a handful of operations.

Worked Example 3.6.2

Insertion Sort's Worst Case

Here's an example from algorithm analysis. In the worst case, insertion sort compares the i-th element against up to i of the elements already sorted. For an array of seven elements, the six passes compare one, two, three, four, five, and six times. That's an arithmetic series with first term one, difference one, and six terms. So the total is six over two, times one plus six, which is twenty-one comparisons. For an array of any size n, the passes contribute one through n minus one comparisons. By the theorem, that's n minus one, times n, over two. That quadratic expression is exactly why insertion sort's worst-case running time grows like n squared. An arithmetic series sits right inside a running-time analysis.

Table 3.6.1

3 Equivalent Approaches

This table compares three equivalent approaches. When the first and last terms are both known, use n over two times the first term plus the last. It takes a fixed number of operations. When only the first term and the common difference are known, use n over two, times two a one plus n minus one d. It substitutes for the last term directly. And adding term by term takes work that grows with n. It's useful for spot-checking a closed-form result, or when n is small.

Closed Form vs Repeated Addition

From a Loop to 3 Operations

Section three point five computed a running total with an accumulator, one addition per term. That cost grows with n. The theorem replaces the loop with one multiplication, one addition, and one division, however large n is. This code computes the same series both ways. The loop adds each term and steps it forward by d. The formula computes the last term, then applies the theorem. With first term five, difference three, and one thousand terms, the last term is three thousand two. So the sum is five hundred times three thousand seven, which is one million five hundred three thousand five hundred. Three operations, and the same answer the thousand-step loop produces.

Takeaways

Pair, Cancel, Collapse

To summarize. Every finite arithmetic series has a closed-form sum, n over two times the first term plus the last. It follows from writing the series forwards and backwards and adding. Every matched pair equals the same constant, the first term plus the last. The equivalent form, in terms of a one and d, is useful when the last term hasn't been computed yet. The argument works identically for even and odd n. And replacing a loop with the formula, turns work that grows with n into a fixed number of operations. Here's something to think about. Pair a five-term series literally, a one with a five, and a two with a four. What happens to the middle term, a three? Work through that case by hand. And explain why the forwards-plus-backwards version never has to answer that question at all.

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FAQ

Questions students ask us

About Sum of Arithmetic Series

What is an arithmetic series, and how is it different from an arithmetic sequence?
An arithmetic series is the sum of the terms of a finite arithmetic sequence, while the sequence is just the list of terms. If a_i = a_1 + (i-1)d, then S_n = a_1 + a_2 + \cdots + a_n is the series. It is also the n-th partial sum. Adding term by term takes n-1 additions.
How do you derive the sum formula for an arithmetic series by writing it forwards and backwards?
Write S_n forwards, write it again backwards, and add the two copies term by term. This gives 2S_n as n pairs a_k + a_{n+1-k}. Each pair equals 2a_1 + (n-1)d = a_1 + a_n, since k cancels. So 2S_n = n(a_1+a_n), and S_n = \frac{n}{2}(a_1+a_n).
Which form of the arithmetic series formula should I use, and how do I apply it?
Use S_n = \frac{n}{2}(a_1+a_n) when the last term is known. Use S_n = \frac{n}{2}\left(2a_1+(n-1)d\right) when you only have a_1, d, and n. For a_1=60, d=25, n=12, the last term is 60+11\cdot 25=335. Then S_{12}=6\cdot 395=2370. The second form gives the same value.
Why use the closed-form formula instead of just adding the terms one by one?
The formula takes a fixed number of operations no matter how large n is, while repeated addition grows with n. For a_1=5, d=3, and 1000 terms, the last term is 5+999\cdot 3=3002. Then S_{1000}=500\cdot 3007=1{,}503{,}500. A 1000-step loop gives the same answer. Direct addition is still useful for checking a result or when n is small.

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