How Young Gauss Discovered the AP Formula

This is my first piece of writing here, and what better way to start than by talking about one of my favorite stories in the mathematical world?

I usually narrate this story to my students in Discrete Math at the Stony Brook University Tutoring Center because I feel it makes seemingly boring mathematics more fun and relatable.

So, let us start the story.

According to a famous mathematical anecdote, the legendary mathematician Carl Friedrich Gauss was in grade 3 at this time. He was in his classroom when the teacher asked all the students to compute the following summation:

S = 1 + 2 + 3 + \cdots + 98 + 99 + 100

The teacher wanted to keep the students busy and quiet for a long time with the help of this problem.

But little did the teacher know that one of the greatest mathematicians to ever exist was sitting in that classroom.

To everyone’s surprise, Gauss was able to compute the sum in mere seconds.

The teacher was shocked and asked Gauss to explain how he had computed it so quickly.

The young Gauss explained:

I realized that the sum of the first and last numbers is $101$ ( $100+1$ ). Similarly, the sum of the second and second-to-last numbers is $101$ ( $2+99$ ). Again, the sum of the third and third-to-last numbers is also $101$ ( $3+98$ ). And so on. This means every such pair sums to $101$ . There are $100$ terms in total, so there are $50$ such pairs. This gives us the sum $50 \times 101 = 5050$ .

The teacher was astonished, flabbergasted, and whatever other adjectives you could think of in this regard.

This led to the birth of the arithmetic progression formula.

Let us generalize the formula.

Let us say we have a sequence of the following form:

a,\ a+d,\ a+2d,\ \ldots,\ a+(n-1)d

Here, $a$ is the first term and $d$ is the common difference.

In the story, you can notice that $a=1$ (the first term is $1$ ) and $d=1$ (the difference between two consecutive terms is $1$ ). Also, $n=100$ because there are $100$ terms.

So, let us compute $S$ using the idea Gauss proposed:

S = a + (a+d) + (a+2d) + \cdots + (a+(n-2)d) + (a+(n-1)d) \tag{i}

Behold!

We can rewrite $S$ as:

S = (a+(n-1)d) + (a+(n-2)d) + \cdots + (a+2d) + (a+d) + a \tag{ii}

Add (i) and (ii):

2S = [2a+(n-1)d] + [2a+(n-1)d] + \cdots + [2a+(n-1)d]

Realize that we are using the same idea: the sums of the pairs are all the same.

Now, since the total number of terms is $n$ , the total number of terms on the right-hand side of the equation is also $n$ .

Rewriting our equation:

2S = n[2a+(n-1)d]

S = \frac{n}{2}[2a+(n-1)d]

And this, ladies and gentlemen, is the generalized formula for the sum of an arithmetic progression.

We can also express it in terms of the first and last terms. If the last term is $l = a + (n-1)d$ , then the formula becomes:

S = \frac{n}{2}(a+l)

In other words, the sum of an arithmetic progression is the number of terms multiplied by the average of the first and last terms.

That is it! The main idea of this article was to share the beauty of mathematics and show that it is not as boring as people might generally think.

To end, here is a quote attributed to Gauss:

“Mathematics is the queen of the sciences, and number theory is the queen of mathematics.”

— Carl Friedrich Gauss