Math Lesson 12.2.2 - The Gauss Method and Arithmetic Series

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Welcome to our Math lesson on The Gauss Method and Arithmetic Series, this is the second lesson of our suite of math lessons covering the topic of Working with Arithmetic and Geometric Series. How to find the Sum of the First n-Terms of a Series., you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

The Gauss Method and Arithmetic Series Explained

As said above, the main objective when dealing with an arithmetic series is to find the sum of the first n terms of the corresponding arithmetic progression (sequence). We solved such an example in the previous lesson, where the sum of the first 5 terms in an arithmetic progression (sequence) was considered. However, that was a specific case. If the calculation of this quantity (the sum of the first n terms) is not generalized through a formula (like those used for finding the general term of a sequence), we are not on the right path. Let's try to find a general formula for the sum of the first n terms of an arithmetic series. The first who discovered a formula for this was the famous scientist Gauss when he was just a kid. He was asked to calculate the sum of the first 100 natural numbers, i.e. 1 + 2 + 3 + … + 98 + 99 + 100. Surprisingly, Gauss found this sum in a very short time by using the following procedure:

In his way, Gauss obtained 50 equal sums where each of them is 101. In this way, he calculated the sum of the first 100 natural numbers by multiplying the above two numbers, i.e.

Since all the above sums are equal to the first term plus the last term (x₁ + x_n), Gauss realized that multiplying (x₁ + x_n) by n/2/, which represents the number of pairs obtained (here we have 50 pairs, as 50 = 1/2 of 100), gives the sum of the first n terms of an arithmetic sequence. In this way, we obtain

Example 2

Use the Gauss Method to find the following sums:

1. 3 + 6 + 9 + … + 147 + 150
2. 4 + 8 + … + 80 + 84

Solution 2

1. We can write
   3 + 6 + 9 + ⋯ + 147 + 150
   = 3 ∙ (1 + 2 + 3 + ⋯ + 49 + 50)
   Thus, for the part inside the brackets we can write x₁ = 1, x_n = 50 and n = 50.
   Hence,
   S_n = 3 ∙ (x₁ + x_n ) ∙ n/2
   = 3 ∙ (1 + 50) ∙ 50/2
   = 3 ∙ 51 ∙ 50/2
   = 3825
2. We can write
   4 + 8 + 12 + ⋯ + 80 + 84
   = 4 ∙ (1 + 2 + 3 + ⋯ + 20 + 21)
   Thus, for the part inside the brackets we can write x₁ = 1, x_n = 21 and n = 21.
   Hence,
   S_n = 4 ∙ (x₁ + x_n ) ∙ n/2
   = 4 ∙ (1 + 21) ∙ 21/2
   = 4 ∙ 22 ∙ 21/2
   = 924

We can use the Gauss Formula in whatever arithmetic series to fit a specific situation. However, first, we must make some arrangements. For example, to find the following sum 7 + 11 + 15 + … + 75 + 79 we do like this:

Since x₁ = 7, x_n = 79 and d = 4, we have from the previous tutorial

In the specific case, we have

Therefore, we obtain for the above sum

Example 3

Calculate the following sum

Solution 3

We have x₁ = 22, x_n = 148 and d = 7.

Moreover, since

we obtain the number of terms n of the given sum by rearranging the above expression

Therefore, using the Gauss formula yields

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7. Continuing learning sequences and series - read our next math tutorial: Binomial Expansion and Coefficients

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