There's a famous (and probably apocryphal) story about the mathematician Carl Friedrich Gauss that goes something like this:

Gauss was 9 years old, and sitting in his math class. He was a genius even at this young age, and as such was incredibly bored in his class and would always goof off and get into trouble. One day his teacher wanted to punish him for goofing off, and told him that if he was so smart, why didn't he go sit in the corner and add up all the integers from 1 to 100? Gauss went and sat in the corner, but didn't pick up his pencil. The teacher confronted him, saying “Carl! Why aren't you working? I suppose you've figured it out already, have you?” Gauss responded with “Yes – it's 5,050.” The teacher didn't believe him and spent the next ten minutes or so adding everything up by hand, only to find that Gauss was right!

So how did Gauss find the answer so fast? What did he see that his teacher didn't? The answer is simple, really – it's all about pattern recognition. Let's look at the problem more closely.

1 + 2 + 3 + 4 + 5 +...+ 95 + 96 + 97 + 98 + 99 + 100 = ?

Now it's true that adding all that up by hand would take forever, but we don't really need to add it all up by hand. Look at this series from each end simultaneously instead of just left to right. You'll see that we can think of this series as a set of pairs of numbers, each of which adds up to 101:

1 + 2 + 3 + 4 + 5 +...+ 95 + 96 + 97 + 98 + 99 + 100 = ?

1 + 100 = 101

2 + 99 = 101

3 + 98 = 101

4 + 97 = 101

5 + 96 = 101

and so on right through to the middle, where:

48 + 53 = 101

49 + 52 = 101

50 + 51 = 101

So we've got a total of 50 pairs, each of which adds up to 101. Since all our chunks are the same size, we can take a shortcut and simply multiply the size of the pair (101) by the number of pairs (50). Which is really easy to do in your head, since it's just (100 x 50) plus (1 x 50), or 5000 + 50 = 5,050.

This reasoning can actually be extrapolated to work with any arithmetic series, and in fact is how we get the formula for the sum of an arithmetic series. Check it out:

Each pair added up to the same number, so we could actually use the mathematical expression for any one of those pairs in our formula. Since the first and last term are the ones most often known, let's use those. Remember, the last term in the series is written as a

The first term (a

Check out the first term in each of our pairs above. They range from 1 to 50, so there are 50 terms – exactly half the number of terms in the series. Which makes sense, since we're pairing up the terms and that by definition gives us half as many pairs as terms.

So 50 in our example is represented by one-half of n, or n/2.

And what are we doing with these two bits of information? Multiplying them together. So our final formula would be:

Σ = (n/2)(a1 + an)

Now, sometimes you see this formula written as n(a1 + an),

2

or the number of terms times the average of the first and last terms. Practically that is exactly the same thing as the way we wrote it first, it's just written a little differently. But they both have the same value, so if it's easier for you to remember it as the average times the number of terms, do so.

This formula works for any arithmetic series, so the next time you come up against one, remember Gauss and his pairs of terms!

Gauss was 9 years old, and sitting in his math class. He was a genius even at this young age, and as such was incredibly bored in his class and would always goof off and get into trouble. One day his teacher wanted to punish him for goofing off, and told him that if he was so smart, why didn't he go sit in the corner and add up all the integers from 1 to 100? Gauss went and sat in the corner, but didn't pick up his pencil. The teacher confronted him, saying “Carl! Why aren't you working? I suppose you've figured it out already, have you?” Gauss responded with “Yes – it's 5,050.” The teacher didn't believe him and spent the next ten minutes or so adding everything up by hand, only to find that Gauss was right!

So how did Gauss find the answer so fast? What did he see that his teacher didn't? The answer is simple, really – it's all about pattern recognition. Let's look at the problem more closely.

1 + 2 + 3 + 4 + 5 +...+ 95 + 96 + 97 + 98 + 99 + 100 = ?

Now it's true that adding all that up by hand would take forever, but we don't really need to add it all up by hand. Look at this series from each end simultaneously instead of just left to right. You'll see that we can think of this series as a set of pairs of numbers, each of which adds up to 101:

1 + 2 + 3 + 4 + 5 +...+ 95 + 96 + 97 + 98 + 99 + 100 = ?

1 + 100 = 101

2 + 99 = 101

3 + 98 = 101

4 + 97 = 101

5 + 96 = 101

and so on right through to the middle, where:

48 + 53 = 101

49 + 52 = 101

50 + 51 = 101

So we've got a total of 50 pairs, each of which adds up to 101. Since all our chunks are the same size, we can take a shortcut and simply multiply the size of the pair (101) by the number of pairs (50). Which is really easy to do in your head, since it's just (100 x 50) plus (1 x 50), or 5000 + 50 = 5,050.

This reasoning can actually be extrapolated to work with any arithmetic series, and in fact is how we get the formula for the sum of an arithmetic series. Check it out:

Each pair added up to the same number, so we could actually use the mathematical expression for any one of those pairs in our formula. Since the first and last term are the ones most often known, let's use those. Remember, the last term in the series is written as a

_{n}, where n is the number of terms in the series. So that 101 will be represented by:The first term (a

_{1}) + the last term (a_{n}), or (a_{1}+ a_{n})Check out the first term in each of our pairs above. They range from 1 to 50, so there are 50 terms – exactly half the number of terms in the series. Which makes sense, since we're pairing up the terms and that by definition gives us half as many pairs as terms.

So 50 in our example is represented by one-half of n, or n/2.

And what are we doing with these two bits of information? Multiplying them together. So our final formula would be:

Σ = (n/2)(a1 + an)

Now, sometimes you see this formula written as n(a1 + an),

2

or the number of terms times the average of the first and last terms. Practically that is exactly the same thing as the way we wrote it first, it's just written a little differently. But they both have the same value, so if it's easier for you to remember it as the average times the number of terms, do so.

This formula works for any arithmetic series, so the next time you come up against one, remember Gauss and his pairs of terms!