Some Special Series (Part 1)

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Arithmetic Series

Back in the late 18^th century, a German school teacher assigned his room full of young students an assignment: "Add up the integers from 1 to 100." It was intended as busy work - something to keep them quiet for a little while. After just a few seconds, one of his students, young Carl Friedrich Gauss raised his hand with the answer. Ten year old Carl, who would one day become of history's most famous mathematicians had seen a pattern. If you write the numbers in order and add together the first and last number, the result is 1 + 100 = 101. If you take the second number and the second to the last and add them together, the result is 2 + 99 = 101.

The young Gauss had realized that, as you worked your way into the center, each pair of numbers gave you the same sum: 101. Because there were 50 pairs in the list, the total had to be 50 · 101 or 5050.

Sum of the First n Integers

$$\sum_{i=1}^n i = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$$

The easy way to remember this is to see that n + 1 is the sum of the first number and the last number, i.e. it's the constant amount that Gauss found by adding together pairs of numbers from the list, and n / 2 is just the number of pairs in the list.

Notice how our sum is based on the terms of an arithmetic sequence. This kind of series has an obvious name.

Arithmetic Series

An arithmetic series is a series formed by adding together the terms of an arithmetic sequence.

Example 1 - Arithmetic Series

Find the sum of the integers from 1 to 1034.

To find this value we can use the formula for the sum of the first n integers with n = 1034.

$$1 + 2 + 3 + ... + 1033 + 1034 = \sum_{i=1}^{1034}i$$ $$1 + 2 + 3 + ... + 1033 + 1034 = \frac{1034(1034+1)}{2} = 535095$$

Example 2 - Arithmetic Series

Find the sum of the integers from 0 to 200.

This question changed it up a little by starting from 0 instead of 1 but adding 0 doesn't change the value of the sum so this is the same as asking for the sum of the integers from 1 to 200.

$$0 + 1 + 2 + ... + 199 + 200 = 1 + 2 + ... + 199 + 200$$ $$0 + 1 + 2 + ... + 199 + 200 = \sum_{i=1}^{200} i$$ $$0 + 1 + 2 + ... + 199 + 200 = \frac{200(200+1)}{2}$$ $$0 + 1 + 2 + ... + 199 + 200 = 20100$$

Example 3 - Arithmetic Series

Find a formula for the sum of the even integers between 2 and 50.

We can do this by getting a little creative with our formula. The sum of the even integers between 2 and 50 looks like

2 + 4 + 6 + ... + 48 + 50

Applying the procedures that we used in the last section, a formula for the i^th term of our series would be 2i. That makes our series

$$2 + 4 + 6 + ... + 48 + 50 = \sum_{i=1}^{25}2i$$

Notice how I only let the series go up to 25. That's because 2 · 25 gives us our top value (50).

We saw in the last section's exercises, that we could take a number in front of the formula in a series and factor it out. That makes our equation

$$2 + 4 + 6 + ... + 48 + 50 = 2\sum_{i=1}^{25}2i$$

But the sum on the right hand side is just the sum of the integers from 1 to 25. Our formula tells us that that equals $\frac{25(25 + 1)}{2} = 325$ so our answer is

$$2 + 4 + 6 + ... + 48 + 50 = 2\cdot325 = 650$$

Take a close look at the parts of that equation. We were looking for the sum of the even integers from 1 to 50 and our formula ended up being, "twice the sum of all the integers between 1 and 25", i.e. "twice the sum of all the integers between 1 and half the maximum number on our list".

Example 4 - Arithmetic Series

Find the sum of the integers from 596 to 835.

Our arithmetic series formula assumes that our series starts at 1 which this one doesn't. To find this value, we're going to need to be a little creative. Suppose we start with the sum of the numbers from 1 to 835:

1 + 2 + 3 + ... + 834 + 835

I can split that into two pieces by taking the part below 596 and the part above 596:

(1 + 2 + ... + 594 + 595) + (596 + 597 + ... + 835) = 1 + 2 + 3 + ... + 834 + 835

Using our formula, we know that the right hand side is

If I replace the right side of our equation with 349030, it becomes

(1 + 2 + ... + 594 + 595) + (596 + 597 + ... + 835) = 349030

Now look at the left hand side. I can also use our formula to calculate 1 + 2 + ... + 595.

If I replace the first parentheses on the left hand side with that value, our equation becomes

177310 + (596 + 597 + ... + 835) = 349030

but if I subtract 177310 from both sides of that equation, I get

596 + 597 + ... + 835 = 171720

The left side of that equation is the sum that we were looking for, i.e. the sum of the numbers from 596 to 835, which means it must equal 171720.

It would be nice to have a formula for the sum of the integers between any two numbers rather than having to start at 1. Fortunately, we can use the same reasoning for this situation as we did with a series that started at 1.

The official formula looks like this

Sum of the Integers from m to n

$$\sum_{i=m}^n = m + (m + 1) + ... + n = \frac{(m - n + 1)(m + n)}{2}$$

Example 5 - Arithmetic Series

Find the sum of the integers from 596 to 835.

This is the same question that we saw in Example 5 but now we can answer it a lot faster using our formula.

$$596 + 597 + ... + 835 = \sum_{i=596}^835 i = \frac{(835 - 596 + 1)(835 + 596)}{2}= \frac{240 \cdot 1431}{2}= 171720$$