# Stay Focused

From Kirk Cousins, quarterback of the Washington Redskins:

Sometimes our guests ask why I have this hanging above my desk. It’s an old high school math quiz when I didn’t study at all and got a C+… just a subtle reminder to me of the importance of preparation. If I don’t prepare I get C’s!

# What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 15

I did not know — until I read Gamma (page 168) — that there actually is a formula for generating $n$th prime number by directly plugging in $n$. The catch is that it’s a mess:

$p_n = 1 + \displaystyle \sum_{m=1}^{2^n} \left[ n^{1/n} \left( \sum_{i=1}^m \cos^2 \left( \pi \frac{(i-1)!+1}{i} \right) \right)^{-1/n} \right]$,

where the outer brackets $[~ ]$ represent the floor function.

This mathematical curiosity has no practical value, as determining the 10th prime number would require computing $1 + 2 + 3 + \dots + 2^{10} = 524,800$ different terms!

When I researching for my series of posts on conditional convergence, especially examples related to the constant $\gamma$, the reference Gamma: Exploring Euler’s Constant by Julian Havil kept popping up. Finally, I decided to splurge for the book, expecting a decent popular account of this number. After all, I’m a professional mathematician, and I took a graduate level class in analytic number theory. In short, I don’t expect to learn a whole lot when reading a popular science book other than perhaps some new pedagogical insights.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

# Useless Numerology for 2016: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post.

Part 1: Introduction.

Part 2: $2016 = 2^{10}+2^9+2^8+2^7+2^6+2^5 = 2^{11} - 2^5$.

Part 3: $2016 = 1 + 2 + \dots + 63$.

Part 4: $2016 = (1 + 2 + \dots + 9)^2 - (1+2)^2 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$

Part 5: $2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2$

# My Mathematical Magic Show: Part 5d

Last March, on Pi Day (March 14, 2015), I put together a mathematical magic show for the Pi Day festivities at our local library, compiling various tricks that I teach to our future secondary teachers. I was expecting an audience of junior-high and high school students but ended up with an audience of elementary school students (and their parents). Still, I thought that this might be of general interest, and so I’ll present these tricks as well as the explanations for these tricks in this series. From start to finish, this mathematical magic show took me about 50-55 minutes to complete. None of the tricks in this routine are original to me; I learned each of these tricks from somebody else.

Though this wasn’t part of my Pi Day magic show, I recently read an interesting variant on my fourth trick. The next time I do a mathematical show, I’ll do this trick next — not to amaze and stun my audience, but to see if my audience can figure out why it works. Each member of the audience will need to have a calculator (a basic four-function calculator will suffice). Here’s the patter:

I want you to take out your calculator. Using only the digits 1 through 9, there are three rows, three columns, and two diagonals. I want you to pick either a row, a column, or a diagonal. Then I want you to enter a three digit number using those numbers. For example, if you chose the first row, you can enter 123 or 312 or 231 or any three-digit number using each digit once.

Now, I want you to multiply this number by another three-digit number. So hit the times button.

(pause)

Now, choose another row, column, or diagonal and type in another three-digit number, using each of the three digits once.

(pause)

Now hit the equals button to multiply those two numbers together.

(pause)

Is everyone done? The product you just computed should have either five or six digits. I want you to concentrate on one of those digits. Just make sure that you concentrate on a digit other than zero, because zero is boring. So concentrate on a nonzero digit.

(pause)

(I point to someone.) Without telling me the digit you chose, please tell me the other digits in your product.

The audience member will say something like, “3, 7, 9, and 2.” To which I’ll reply in three seconds or less, “The number you chose was 6.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “1, 1, 9, 7, and 2.” I’ll answer, “The number you chose was 7.”

And then I’ll repeat this a few times, and everyone’s amazed that I knew the different numbers that were chosen.

Clearly this works using the same logic as my fourth magic trick: the product is always a multiple of 9, and so I can add the digits to figure out the missing digit. The more interesting question is: Why is the product always a multiple of 9?

This works because each of the factors of the product is a multiple of 3. Let’s take another look at the calculator.

If the first row is chosen, the sum of the digits is 1+2+3 = 6, a multiple of 3. And it doesn’t matter if the number is 123 or 312 or 231… the order of the digits is unimportant.

If the second row is chosen, the sum of the digits is 4+5+6 = 15, a multiple of 3.

If the third row is chosen, the sum of the digits is 7+8+9 = 24, a multiple of 3.

If the first column is chosen the sum of the digits is 1+4+7=12, a multiple of 3.

If the second column is chosen, the sum of the digits is 2+5+8 = 15, a multiple of 3.

If the third column is chosen, the sum of the digits is 3+6+9 = 18, a multiple of 3.

If one diagonal is chosen, the sum of the digits is 1+5+9 = 15, a multiple of 3.

If the other diagonal is chosen, the sum of the digits is 3+5+7 = 15, a multiple of 3.

This can be stated more succinctly using algebra. The digits in each row, column, and diagonal form an arithmetic sequence. For each row, the common difference is 1. For each column, the common difference is 3. And for a diagonal, the common difference is either 2 or 4. If I let $a$ be the first term in the sequence and let $d$ be the common difference, then the three digits are $a$, $a + d$, and $a + 2d$, and their sum is

$a + (a+d) + (a+ 2d) = 3a + 3d = 3(a+d)$,

which is a multiple of 3. (Indeed, the sum is 3 times the middle number.)

So each factor is a multiple of 3. That means the product has to be a multiple of $9$. In other words, if the first factor is $3m$ and the second factor is $3n$, where $m$ and $n$ are integers, their product is equal to

$(3m)(3n) = 9(mn)$,

which is clearly a multiple of 9. Therefore, I can use the same adding-the-digits trick to identify the missing digit.

# Arithmetic and Geometric Series: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on how I remind students about Taylor series. I often use this series in a class like Differential Equations, when Taylor series are needed but my class has simply forgotten about what a Taylor series is and why it’s important.

Part 1: Deriving the formulas for the $n$th term of arithmetic and geometric sequences.

Part 2: Pedagogical thoughts on conceptual barriers that students often face when encountering sequences and series.

Part 3: The story of how young Carl Frederich Gauss, at age 10, figured out how to add the integers from 1 to 100 in his head.

Part 4: Deriving the formula for an arithmetic series.

Part 5: Deriving the formula for an arithmetic series, using mathematical induction. Also, extensions to other series.

Part 6: Deriving the formula for an arithmetic series, using telescoping series. Also, extensions to other series.

Part 7: Pedagogical thoughts on assessing students’ depth of understanding the formula for an arithmetic series.

Part 8: Deriving the formula for a finite geometric series.

Part 9: Infinite geometric series and Xeno’s paradox.

Part 10: Deriving the formula for an infinite geometric series.

Part 11: Applications of infinite geometric series in future mathematics courses.

Part 12: Other commonly-arising infinite series.

# Formula for an arithmetic series (Part 7)

As we’ve discussed, the formula for an arithmetic series is

$S_n = \displaystyle \frac{n}{2} (2a_1 + [n-1] d) = \displaystyle \frac{n}{2} (a_1 + a_n)$,

where $n$ is the number of terms, $a_1$ is the first term, $d$ is the common difference, and $a_n$ is the last term. This formula may be more formally expressed as

$S = \displaystyle \sum_{k=1}^n a_k = \displaystyle \frac{n}{2} (2a_1 + [n-1] d) = \displaystyle \frac{n}{2} (a_1 + a_n)$

For homework and on tests, students are asked to directly plug into this formula and to apply this problem with word problems, like finding the total number of seats in an auditorium with 50 rows, where there are 12 seats in the front row and each row has two more seats than the row in front of it.

In my opinion, the ability to solve questions like the one below is the acid test for determining whether a student — who I assume can solve routine word problems like the one above — really understands series or is just familiar with series. In other words, if a student can solve routine word problems but is unable to handle a problem like the one below, then there’s still room for that student’s knowledge of series to deepen.

Calculate $\displaystyle \sum_{k=11}^{60} (5k - 2)$

There are two reasonable approaches for solving this problem.

Solution #1. Notice that $5k - 2 = 5(k-1) + 5 - 2 = 3 + 5(k-1)$. So this is really an arithmetic series whose first term is $3$ and whose common difference is $5$. Therefore,

$S = \displaystyle \sum_{k=1}^{60} a_k = \displaystyle \frac{60}{2} (2[3] + [60-1] 5)=9030$

However, I’m supposed to start the series on $k=11$, not $k=1$. That means that I need to subtract off the first ten terms of the above series. Now

$S = \displaystyle \sum_{k=1}^{10} a_k = \displaystyle \frac{10}{2} (2[3] + [10-1] 5)= 255$

Finally,

$\displaystyle \sum_{k=11}^{60} a_k = \displaystyle \sum_{k=1}^{60} a_k - \displaystyle \sum_{k=1}^{10} a_k = 9030 - 255 = 8775$

Solution #2. Writing out the terms, we see that

$\displaystyle \sum_{k=11}^{60} (5k - 2) = (5[11]-2) + (5[12]-2) + \dots + (5[60]-2)$

or

$\displaystyle \sum_{k=11}^{60} (5k - 2) = 53+58 + \dots +298$

The right-hand side is an arithmetic series whose “first” term is $53$ and whose last ($50$th) term is $298$. Therefore,

$\displaystyle \sum_{k=11}^{60} (5k - 2) = \frac{50}{2} (53+298) = 8775$

Of the two solutions, I suppose I have a mild preference for the first, as the second solution won’t work for something like $\displaystyle \sum_{k=11}^{60} k^2$. However, both solution demonstrate that the student is actually thinking about the meaning of the series instead of just plugging numbers in a formula, and so I’d be happy with either one in a Precalculus class.

# Formula for an arithmetic series (Part 6)

In the previous posts of this series, I described two methods of deriving the formula

$\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$

The first method concerned reversing the terms of the sum (or, almost equivalently, taking the terms in pairs). The second method used mathematical induction.

Mathematical induction can be applied to arithmetic series as well as other series. However, the catch is that you have to know the answer before proving that the answer actually is correct. By contrast, the first method did not require us to know the answer in advance — it just fell out of the calculation — but it cannot be applied to series that are not arithmetic.

Here’s a third method using the principle of telescoping series. This method has the strengths of the previous two methods: it does not require us to know the answer in advance, and it can also be applied to some other series which are not arithmetic.

To begin, consider the sum

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2]$

At this early point, students often object, “Where did that come from?” I’ve said it before but I’ll say it again: I tell them my usual tongue-in-cheek story that this idea comes from the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

In any event, I will evaluate this sum in two different ways.

Step 1. Just write out the terms of the series, starting from $k=1$ and ending with $k =n$.

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = [1^2 - 0^2] + [2^2 - 1^2] + [3^2 - 2^2] + \dots + [n^2 - (n-1)^2]$

Notice that, on the right-hand side, the $1^2$ terms cancel, the $2^2$ terms cancel, and so on. In fact, almost everything cancels. The only two terms that aren’t cancelled are the $0^2$ and $n^2$ terms. Therefore,

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = n^2 - 0^2 = n^2$

Step 2. Next, we’ll rewrite the original sum by expanding out the terms inside of the sum:

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n [k^2 - (k^2 -2k + 1)]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n [2k-1]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n 2k - \displaystyle \sum_{k=1}^n 1$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 2\displaystyle \sum_{k=1}^n k - \displaystyle \sum_{k=1}^n 1$

Step 3. Of course, these different looking answers from Steps 1 and 2 have to be the same, so let’s set them equal to each other:

$2\displaystyle \sum_{k=1}^n k - \displaystyle \sum_{k=1}^n 1 = n^2$

There is one unknown in this equation, $\displaystyle \sum_{k=1}^n k$. The second sum is just the constant $1$ added to itself $n$ times, and so $\displaystyle \sum_{k=1}^n 1 = n$. Therefore, we solve for the unknown:

$2 \left(\displaystyle \sum_{k=1}^n k \right) - n = n^2$

$2 \left(\displaystyle \sum_{k=1}^n k \right) = n^2 + n$

$\displaystyle \sum_{k=1}^n k = \displaystyle \frac{n^2 + n}{2}$

The beauty of this approach is that this approach can be continued. For example, to obtain $\displaystyle \sum_{k=1}^n k^2$, we begin with

$\displaystyle \sum_{k=1}^n [k^3 - (k-1)^3]$

Step 1. By telescoping series,

$\displaystyle \sum_{k=1}^n [k^3 - (k-1)^3] = n^3 - 0^3 = n^3$

Step 2. Using the binomial theorem,

$\displaystyle \sum_{k=1}^n [k^3 - (k-1)^3] = \displaystyle \sum_{k=1}^n [k^3 - (k^3 -3k^2+3k- 1)]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 3\displaystyle \sum_{k=1}^n k^2 - 3\displaystyle \sum_{k=1}^n k + \displaystyle \sum_{k=1}^n 1$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 3\displaystyle \sum_{k=1}^n k^2 - 3\left( \frac{n(n+1)}{2} \right) + n$

Step 3. Setting these two expressions equal to each other,

$3\displaystyle \sum_{k=1}^n k^2 - 3\left( \frac{n(n+1)}{2} \right) + n= n^3$

And we eventually conclude that:

$\displaystyle \sum_{k=1}^n k^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

And then this could be continued to obtain closed-form expressions for higher exponents of $k$.

# Formula for an arithmetic series (Part 5)

In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student’s first example of a proof by mathematical induction. Recall, from Wikipedia:

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers.

The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

1. The basis (base case): prove that the statement holds for the first natural number $n$. Usually, $n=0$ or $n=1$.
2. The inductive step: prove that, if the statement holds for some natural number $n$, then the statement holds for $n+1$.

The hypothesis in the inductive step that the statement holds for some $n$ is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for $n+1$.

As an inference rule, mathematical induction can be justified as follows. Having proven the base case and the inductive step, then any value can be obtained by performing the inductive step repeatedly. It may be helpful to think of the domino effect. Consider a half line of dominoes each standing on end, and extending infinitely to the right. Suppose that:

1. The first domino falls right.
2. If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right.

With these assumptions one can conclude (using mathematical induction) that all of the dominoes will fall right.

Mathematical induction… works because $n$ is used to represent an arbitrary natural number. Then, using the inductive hypothesis, i.e. that $P(n)$ is true, show $P(k+1)$ is also true. This allows us to “carry” the fact that $P(0)$ is true to the fact that $P(1)$ is also true, and carry $P(1)$ to $P(2)$, etc., thus proving $P(n)$ holds for every natural number $n$.

When students first encounter mathematical induction (in either Precalculus, Discrete Mathematics, or Real Analysis), the theorems that students are asked to prove usually fall into four categories:

1. Calculating a series (examples below).
2. Statements concerning divisibility (for example, proving that $4$ is always a factor of $5^n-1$).
3. Finding a closed-form expression for a recursively defined sequence (for example, if $a_1 = 4$ and $a_n = 3a_{n-1}$ if $n \ge 2$, proving that $a_n = 4 \times 3^{n-1}$0
4. Statements concerning inequality (for example, proving that $n! > 4^n$ if $n \ge 9$0

Here’s a common first example of mathematical induction applied to an arithmetic series. Notice that the statement of the theorem matches the form $\displaystyle \frac{n}{2}(a_1 + a_n)$ seen earlier in this series (pardon the pun) of posts.

Theorem. $1 + 2 + \dots + (n-1) + n = \displaystyle \frac{n(n+1)}{2}$

Proof. Induction on $n$.

$n = 1$: The left-hand is simply $1$, while the right-hand side is $\displaystyle \frac{(1)(2)}{2}$, which is also equal to $1$. So the base case works.

$n$: Assume that the statement holds true for the integer $n$.

$n+1$. If I replace $n$ by $n+1$ in the statement of the theorem, then the right-hand side becomes

$\displaystyle \frac{(n+1)[(n+1)+1]}{2} = \displaystyle \frac{(n+1)(n+2)}{2}$

I find it helpful to describe this to students as my target. In other words, as I manipulate the left-hand side, my ultimate goal is to end up with this target. Once I have done that, then I have completed the proof.

If I replace $n$ by $n+1$ in the statement of the theorem, then the left-hand side will now end on $n+1$ instead of $n$:

$1 + 2 + \dots + (n-1) + n + (n+1)$

Notice that we’ve seen almost all of this before, except for the extra term $n+1$. So we will substitute using the induction hypothesis, carrying the extra $n+1$ along for the ride.

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1)}{2} + (n+1)$

Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target. Most students are completely comfortable doing this, although they typically multiply out the term $n(n+1)$ unnecessarily. Indeed, many early proofs by induction are simplified by factoring out terms whenever possible — in the example below, $(n+1)$ is factored on the last step — as opposed to multiplying them out. In my experience, proofs by induction often serve as a stringent test of students’ algebra skills as opposed to their skills in abstract reasoning.

In any event, here’s the end of the proof:

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1)}{2} + (n+1)$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1) + 2(n+1)}{2}$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(n+1)(n + 2)}{2}$

Mathematical induction can be used to verify formulas for series which are not arithmetic, like

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

$1^3 + 2^3 + \dots + (n-1)^3 + n^3 = \displaystyle \frac{n^2(n+1)^2}{4}$

However, the downside of a proof by induction lies in the word verify, as it’s necessary to actually know what’s going to work before proceeding with the proof.

In the next post, I’ll describe a method of obtaining these series that does not require mathematical induction.

# Formula for an arithmetic series (Part 4)

As I’ve said before, I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

While I’m not a fan of making students memorize formulas, I am a fan of teaching students how to derive formulas. Speaking for myself, if I ever need to use a formula that I know exists but have long since forgotten, the ability to derive the formula allows me to get it again.

Which leads me to today’s post: the derivation of the formulas for the sum of an arithmetic series. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

To get the idea across, consider the arithmetic series

$S = 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 + 43$

Now write the sum in reverse order. This doesn’t change the value of the sum, and so:

$S = 43 + 40 + 37 +34+ 31 + 28 + 25 + 22 + 19 + 16$

Now add these two lines vertically. Notice that $16 + 43 = 59$, $19 + 40 = 59$, and in fact each pair of numbers adds to $59$. So

$2S = 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59$

$2S = 59 \times 10 = 590$

$S = 295$

Naturally, this can be directly confirmed with a calculator by just adding the 10 numbers.

When I show this to my students, they often complain that there’s no way on earth that they would have thought of that for themselves. They wouldn’t have thought to set the sum equal to $S$, and they certainly would not have thought to reverse the terms in the sum. To comfort them, I tell them my usual tongue-in-cheek story that this idea comes from the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

The derivation of the general formula proceeds using the same idea. If $a_1, \dots, a_n$ are the first $n$ terms of an arithmetic sequence, let

$S = a_1 + a_2 + \dots + a_{n-1} + a_n$

Recalling the formula for an arithmetic sequence, we know that

$a_2 = a_1 + d$

$\vdots$

$a_{n-1} = a_1 + (n-2)d$

$a_n = a_1 + (n-1)d$

Substituting, we find

$S = a_1 + [a_1 + d] + \dots + [a_1 + (n-2)d] + [a_1 + (n-1)d]$

As above, we now return the order…

$S = [a_1 + (n-1)d] + [a_1 + (n-2)d] + \dots + [a_1 + d] + a_1$

… and add the two equations:

$2S = [2a_1 + (n-1)d] + [2a_1 + d+(n-2)d] + \dots + [2a_1 +(n-2)d+ d] + [2a_1+(n-1)d]$

$2S = [2a_1 + (n-1)d] + [2a_1 + (n-1)d] + \dots + [2a_1 +(n-1)d] + [2a_1+(n-1)d]$

$2S = n[2a_1 + (n-1)d]$

$S = \displaystyle \frac{n}{2} [2a_1 + (n-1)d]$

We also note that the formula may be rewritten as

$S = \displaystyle \frac{n}{2} [a_1 + \{a_1 + (n-1)d\} ]$

or

$S = \displaystyle \frac{n}{2} [a_1 + a_n]$

This latter form isn’t too difficult to state as a sentence: the sum of a series with $n$  is the average of the first and last terms, multiplied by the number of terms.

Indeed, I have seen textbooks offer proofs of this formula by using the same logic that young Gauss used to find the sum $1 + 2 + \dots + 99 + 100$. The “proof” goes like this: Take the terms in pairs. The first term plus the last term is $a_1 + a_n$. The second term plus the second-to-last term is $a_2 + a_{n-1} = a_1 + d + a_n - d = a_1 + a_n$. And so on. So each pair adds to $a_1 + a_n$. Since there are $n$ terms, there are $n/2$ pairs, and so we derive the above formula for $S$.

You’ll notice I put “proof” in quotation marks. There’s a slight catch with the above logic: it only works if $n$ is an even number. If $n$ is odd, the result is still correct, but the logic to get the result is slightly different. That’s why I don’t particularly recommend using the above paragraph to prove this formula for students, even though it fits nicely with the almost unforgettable Gauss story.

That said, for talented students looking for a challenge, I would recommend showing this idea, then point out the flaw in the argument, and then ask the students to come up with an alternate proof for handling odd values of $n$.

# Calculation of a famous arithmetic series (Part 3)

In this post, we’ll consider the calculation of a very famous arithmetic series… not because the series is particularly important, but because it’s part of a legendary story about one of the greatest mathematicians who ever lived. My frank opinion is that every math teacher should know this story. While I’m not 100% certain about small details of the story — like whether young Gauss was 9 or 10 years old when the following event happened — I’m just going to go with the story as told by the website http://www.math.wichita.edu/history/men/gauss.html.

Carl Friedrich Gauss (1777-1855) is considered to be the greatest German mathematician of the nineteenth century. His discoveries and writings influenced and left a lasting mark in the areas of number theory, astronomy, geodesy, and physics, particularly the study of electromagnetism.

Gauss was born in Brunswick, Germany, on April 30, 1777, to poor, working-class parents. His father labored as a gardner and brick-layer and was regarded as an upright, honest man. However, he was a harsh parent who discouraged his young son from attending school, with expectations that he would follow one of the family trades. Luckily, Gauss’ mother and uncle, Friedrich, recognized Carl’s genius early on and knew that he must develop this gifted intelligence with education.

While in arithmetic class, at the age of ten, Gauss exhibited his skills as a math prodigy when the stern schoolmaster gave the following assignment: “Write down all the whole numbers from $1$ to $100$ and add up their sum.” When each student finished, he was to bring his slate forward and place it on the schoolmaster’s desk, one on top of the other. The teacher expected the beginner’s class to take a good while to finish this exercise. But in a few seconds, to his teacher’s surprise, Carl proceeded to the front of the room and placed his slate on the desk. Much later the other students handed in their slates.

At the end of the classtime, the results were examined, with most of them wrong. But when the schoolmaster looked at Carl’s slate, he was astounded to see only one number: $5050$. Carl then had to explain to his teacher that he found the result because he could see that, $1+100=101$, $2+99=101$, $3+98=101$, so that he could find $50$ pairs of numbers that each add up to $101$. Thus, $50$ times $101$ will equal $5050$.