Pascal’s Triangle and a British game show

So this happened on the popular British game show “University Challenge” on Monday, April 2. This game show pits teams of four from various British universities and is a severe test of the breadth and depth of their knowledge of many fields, including mathematics. A contestant’s response to one math question, asking for the seventh row of Pascal’s triangle, took the UK by storm this week (start at the 26:42 mark of the video below).

Twitter immediately went ablaze. Amazingly, a write-up of this encounter made it into the Times of London, one of the world’s most venerated newspapers (as opposed to the tawdry English tabloids). The above link requires a subscription; here’s a photo of page 13 from the April 4 edition:

I must admit that I’m a little amused by the amount of press that this little encounter received. When I was a kid, I memorized the first few rows of Pascal’s triangle simply from working with it so often, so when a family member told me about this story earlier this week, I knew the answer to the question instantly. I suspect that’s exactly what the contestant did here. (Whether I could have gotten the answer right under the pressure of a quiz show and a national TV audience, on the other hand, is another matter entirely.)

I have a theory as to why this appeared to be a mighty feat of mental arithmetic. The audience may have thought that he was adding the numbers quickly, but I’m guessing that the real purpose of the introductory clause “If 1,1 is the second row of Pascal’s triangle…” is to label that row as the second row instead of the first row (following the usual convention of starting the row and column counts with 0.)

My Favorite One-Liners: Part 91

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Everyone once in a while, a student might make a careless mistake — or just choose an incorrect course of action — that changes what was supposed to be a simple problem into an incredibly difficult problem. For example, here’s a problem that might arise in Calculus I:

Find $f'(x)$ if $f(x) = \displaystyle \int_0^x (1+t^2)^{10} \, dt$

The easy way to do this problem, requiring about 15 seconds to complete, is to use the Fundamental Theorem of Calculus. The hard way is by multiplying out $(1+t^2)^{10}$ — preferably using Pascal’s triangle — taking the integral term-by-term, and then taking the derivative of the result. Naturally, a student who doesn’t see the easy way of doing the problem might get incredibly frustrated by the laborious calculations.

So here’s the advice that I give my students to trying to discourage them from following such rabbit trails:

If you find yourself stuck on what seems to be an incredibly difficult problem, you should ask yourself, “Just how evil do I think my professor is?”

Lessons from teaching gifted elementary students (Part 8g)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received, in the students’ original handwriting. They wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$= \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

$= 110 \times 2^9 + 33 \times 2^{10} + 2^{11}$

$= 92,160$ .

I’m almost done… except my students wanted me to find the square root of this number without using a calculator.

There are a couple ways to do this; the method I chose was directly extracting the square root by hand… a skill that was taught to children in previous generations but has fallen out of pedagogical disfavor with the advent of handheld calculators. I lost my original work, but it would have looked something like this (see the above website for details on why this works):

And so I gave my students their answer: $x \approx 303.578\dots$

Lessons from teaching gifted elementary students (Part 8f)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$= \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

To numerically evaluate $y$ , I use the identity

$\sum_{r=0}^n {n \choose r} = 2^n$ ;

this identity can be proven by using the binomial theorem

$\sum_{r=0}^n {n \choose r} x^r y^{n-r} = (x+y)^n$

and then plugging in $x = 1$ and $y = 1$ . Using this identity, I conclude that

$y = 110 \times 2^9 + 33 \times 2^{10} + 2^{11}$

$= 55 \times 2 \times 2^9 + 33 \times 2^{10} + 2 \times 2^{10}$

$= 55 \times 2^{10} + 33 \times 2^{10} + 2 \times 2^{10}$

$= (55+33+2) \times 2^{10}$

$= 90 \times 2^{10}$ .

Since I know that $2^{10} = 1024$ , it’s now a simple matter of multiplication:

$y = 90 \times 1024 = 92,160$ .

(Trust me; after I showed my students this answer about five minutes after it was posed, I was ecstatic when I confirmed this answer with Mathematica.)

Lessons from teaching gifted elementary students (Part 8e)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

In the first series, I’ll rewrite $11!$ as $11 \times 10 \times 9!$ . Also, in the second series, I’ll rewrite $11!$ as $11 \times 10!$ . Therefore,

$y = \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$y = \displaystyle 110 \sum_{i=0}^{9} \frac{9!}{i!(9-i)!} + 33 \sum_{j=0}^{10} \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

We now see that binomial coefficients appear in each of these series:

$y = \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

I’ll conclude the evaluation of $y$ in tomorrow’s post.

Lessons from teaching gifted elementary students (Part 8d)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

I now use the definition of the binomial coefficient:

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

Since $k! = k(k-1) \times (k-2)!$ and $k! = k \times (k-1)!$ , this simplifies as

$y = \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!}+ 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

In the first series, I’ll use the change of index $i = k-2$ , so that $k = i+2$ and $11-k = 11-(i+2) = 9-i$ . Also, in the first series, the index will change from $k = 2$ to $k = 11$ to $i = 0$ to $i = 9$ .

In the second series, I’ll use the change of index $j = k-1$ , so that $k = j+1$ and $11-k = 11-(j+1) = 10-j$ . Also, in the first series, the index will change from $k = 1$ to $k = 11$ to $j = 0$ to $j = 10$ .

With these changes, I obtain

$y = \displaystyle \sum_{i=0}^{9}\frac{11!}{(i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

I’ll continue the simplification of these series in tomorrow’s post.

Lessons from teaching gifted elementary students (Part 8c)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

In yesterday’s post, I explained how Pascal’s triangle can be used to conclude

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$ ,

thus allowing me to get the top number without getting all of the intermediate steps.

To compute this sum without a calculator, I’ll start rearranging the terms. The reasons for rearranging the terms in this way will become evident later.

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=0}^{11} (k^2 + 2k + 1) {11 \choose k}$

$= \displaystyle \sum_{k=0}^{11} ([k^2 -k] + 3k + 1) {11 \choose k}$

$=\displaystyle \sum_{k=0}^{11} [k(k-1) + 3k + 1] {11 \choose k}$

$= \displaystyle \sum_{k=0}^{11} k(k-1) {11 \choose k} + \sum_{k=0}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

The terms of the first sum are clearly equal to 0 when $k = 0$ and $k =1$ . Also, the $k=0$ term of the second sum is clearly 0. Therefore,

$y = \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

It doesn’t look like I’ve improved matters much with this rearrangement of $y$ ; I’ll continue the solution in tomorrow’s post.

Lessons from teaching gifted elementary students (Part 8b)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

Here’s how I started the problem, using a trick that I use in my mathematical magic show. Suppose that there are only six numbers instead of twelve, and let the six numbers be $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ . Then here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

In other words, the top number can be obtained by using the numbers on the fifth row of Pascal’s triangle (recall that the fifth row of Pascal’s triangle has six numbers on it). Specifically, if I multiply the bottom numbers by the corresponding number in a row of Pascal’s triangle and add them up, I’ll get the number on top without having to compute all of the intermediate steps.

For the problem my students gave me, the bottom row has 12 numbers, which means I’ll need to use the 11th row of Pascal’s triangle. Also, as we’ll see, I was fortunate that my students gave me a simple pattern of consecutive squares for the numbers on the bottom row. Since the numbering in Pascal’s triangle starts on zero, the numbers in the bottom row are $(k+1)^2$ as $k$ varies from 0 to 11.

Putting all this together, I can conclude that

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$ .

Beginning with tomorrow’s post, I’ll discuss how I computed this sum without a calculator.

Lessons from teaching gifted elementary school students (Part 8a)

Here’s a question I once received, in the students’ original handwriting:

Here’s the explanation that my students told me (but didn’t write down): they wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. For example, the lower-left portion of the triangle would build like this (since 1+4=5, 4+9=13, 9+16=25, etc.):

18 38

5 13 25

1 4 9 16

Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

And I produced the answer in less than five minutes.

I’ll reveal how I got the answer so quickly in this series. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own.

Formula for an infinite geometric series (Part 11)

Many math majors don’t have immediate recall of the formula for an infinite geometric series. They often can remember that there is a formula, but they can’t recollect the details. While it’s I think it’s OK that they don’t have the formula memorized, I think is a real shame that they’re also unaware of where the formula comes from and hence are unable to rederive the formula if they’ve forgotten it.

In this post, I’d like to give some thoughts about why the formula for an infinite geometric series is important for other areas of mathematics besides Precalculus. (There may be others, but here’s what I can think of in one sitting.)

1. An infinite geometric series is actually a special case of a Taylor series. (See https://meangreenmath.com/2013/07/05/reminding-students-about-taylor-series-part-5/ for details.) Therefore, it would be wonderful if students learning Taylor series in Calculus II could be able to relate the new topic (Taylor series) to their previous knowledge (infinite geometric series) which they had already seen in Precalculus.

2. An infinite geometric series is also a special case of the binomial series $(1+x)^n$ , when $n$ does not have to be a positive integer and hence Pascal’s triangle cannot be used to find the expansion.

3. Infinite geometric series is a rare case when an infinite sum can be found exactly. In Calculus II, a whole battery of tests (e.g., the Root Test, the Ratio Test, the Limit Comparison Test) are introduced to determine whether a series converges or not. In other words, these tests only determine if an answer exists, without determining what the answer actually is.

Throughout the entire undergraduate curriculum, I’m aware of only four types of series that can actually be evaluated exactly.

An infinite geometric series with $-1 < r < 1$
The Taylor series of a real analytic function. (Of course, an infinite geometric series is a special case of a Taylor series.)
A telescoping series. For example, using partial fractions and cancelling a bunch of terms, we find that

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

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

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

An infinite series that arises from Parseval’s theorem in Fourier analysis.

4. Infinite geometric series are essential for proving basic facts about decimal representations that we often take for granted.

We can justify why $0.\overline{9} = 1$ . See https://meangreenmath.com/2013/09/03/why-does-0-999-1-part-3/ for details.
Along with the Direct Comparison Test, we can be used to justify why a decimal representation $0.d_1d_2d_3\dots$ actually converges and must correspond to a real number. See https://meangreenmath.com/2013/09/01/why-does-0-999-1-part-1/ for details.
We can prove that a fraction of the form $\displaystyle \frac{M}{10^k-1}$ must have a repeating block of length $k$ which contains the digits of $M$ . See https://meangreenmath.com/2013/08/22/thoughts-on-17-and-other-rational-numbers-part-5/ for details.
Going the other direction, we can also convert an repeating decimal representation into a fraction. See https://meangreenmath.com/2013/08/21/thoughts-on-17-and-other-rational-numbers-part-4/ for details.
We can also find representations in other bases besides base 10. See https://meangreenmath.com/2013/08/17/calculator-errors-when-close-isnt-close-enough-part-2 for an example of a repeating binary representation.

5. Properties of an infinite geometric series are needed to find the mean and standard deviation of a geometric random variable, which is used to predict the number of independent trials needed before an event happens. This is used for analyzing the coupon collector’s problem, among other applications.