Engaging students: Using Pascal’s triangle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Lisa Sun. Her topic, from Precalculus: using Pascal’s triangle.

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How could you as a teacher create activity or project that involves your topic?

To introduce Pascal’s Triangle, I would create an activity where it involves coin tossing. I want to introduce them with coin tossing first before bringing in binomial expansions (or any other uses) because coin tossing are much more familiar to majority, if not all, students. Pascal’s Triangle can show you the probability of any combination of coin tossing (aka binomial distribution). Below are a few of the results and how they compare with Pascal’s Triangle:

Afterwards, I would ask the students guiding questions if they see anything interesting about the numbers that we gathered. I want them to notice that each number is the numbers directly above it added together (Ex: 1 + 2 = 3) and how those three numbers form a triangle hence, Pascal’s Triangle.

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B2: How does this topic extend what your students should have learned in previous courses?

In previous courses, students should have already learned about binomial expansions. (Ex: (a+b)2 = a2+ 2ab + b2). This topic extends their prior knowledge even further because Pascal’s Triangle displays the coefficients in binomial expansions. Below are a few examples in comparison with Pascal’s Triangle:

If any of the students are having difficulties expanding any of the binomials or remembering the formula, they can remember Pascal’s Triangle. Using the Pascal’s Triangle for solving binomial expansions can aid the students when it comes to being in a stressful environment (ex: taking a test). Making a connection between their prior knowledge on binomial expansion and Pascal’s Triangle, I believe it would give the students a deeper understanding as to how Pascal’s Triangle was formed.

 

 

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C2: How has this topic appeared in high culture?

There’s a computer scientist, John Biles, at Rochester University in New York State who used the series of Fibonacci numbers to make a piece of music. How do the Fibonacci numbers relate to Pascal’s Triangle you ask? Well, observe the following:

As you can see, the sum of the numbers diagonally gives you the Fibonacci numbers (a series of numbers in which each number is the sum of the two preceding numbers).

John Biles composed a piece called PGA -1 which is based on a Fibonacci sequence. Note that on a piano, from middle C to a one octave C, there are a total of eight white keys (a Fibonacci number). Also, when you do a chromatic C scale which includes all the black keys, there are a total of five black keys (another Fibonacci number) which are also separated in a group of two and three black keys (see the pattern?). When you’re creating chords, let’s take the C chord for example, it consists of the notes C, E, and G. Notice that harmonizing notes are coming from the third note and the fifth note of the whole C scale. So following similar ideas on the use of these numbers/sequences, John Biles was able to compose music.

Here is John Biles full article: http://igm.rit.edu/~jabics//Fibo98/

Here is his composed song: http://igm.rit.edu/~jabics//Fibo98/PGA-1.mp3

The following may be a bit extra, but I also want to include this youtube link of this blogger who was very precise and compared the sequences to current pop music:

[I found this to be super interesting!]

How have different cultures throughout time used this topic in their society?

Hundreds of years before Blaise Pascal (mathematician whom Pascal’s Triangle was named after), many mathematicians in different societies applied their knowledge of the Triangle.

Indian mathematicians used the array of numbers to represent short and long sounds in poetic meters in their chants and conversations. A Chinese mathematician, Chu Shih Chieh, used the triangle for binomial expansions. Music composers, like Mozart and Debussy, used the sequence to compose their music to guide them what notes to play that would be pleasing to the audience. In the past, arithmetic composing was frowned upon however contemporary music to this day is now filled with them. When Pascal’s work on the triangle was published, society began to apply the knowledge of the Triangle towards gambling with dice. In the end, all cultures began to use Pascal’s Triangle similarly in their daily lives.

How can technology be used to effectively engage students with this topic?

The Youtube video above is a great tool for students who are visual learners. This video is to the point and clear with the message as to what Pascal’s Triangle is, the uses of it, and who aided in the discovery of it. I also believe the characters that were being used in this video would be appealing to students. This video was filled with facts that I want my students to know therefore, I would like them to follow along and write down important facts about Pascal’s Triangle. I would like to conclude that technology can be a “force multiplier” for all teachers in their classroom. Instead of having the teacher being the only source of help in a classroom, students can access web site, online tutorials, and more to assist them. What’s great is that students can access this at any time. Therefore, they can re-watch this video again once they’re home when they need a refresher or didn’t understand something the first time.

 

References:

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibInArt.html#othermusic

http://www.mathsisfun.com/pascals-triangle.html

http://ualr.edu/lasmoller/pascalstriangle.html

 

Engaging students: Using Pascal’s triangle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Daniel Herfeldt. His topic, from Precalculus: using Pascal’s triangle.

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A great activity for Pascal’s triangle would be to first have the students find a pattern of odds and evens. The first thing that you would do is to print out blank Pascal’s triangle. You would give each student a paper for them to fill out. They would have to first fill out the triangle themselves. This would give them practice on which numbers to add as well as further see a pattern of what the next one would potentially look like. After they finish, they would have to color in all of the odd numbers a certain color, and followed by coloring all of the even ones a different color. From here, they will see that once you color it is, the even numbers will make an upside down triangle. Next to the biggest triangles, you will see smaller triangles. An example is shown below. When the students have finished, you will show them why it is like that. Then explain what the name of the colored triangle is, which is called the Sierpinski Triangle.

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Pascal’s Triangle is used all over mathematics. It is mainly recognized as how to find the coefficients of binomials, as well as a lot of other uses for binomials. What students and many other people do not know, is that this triangle can be used for much more. For example, you are able to use Pascal’s triangle to find the Fibonacci sequence. Although it may be a little harder to find than the coefficients of binomials, it is still possible. If you add up the numbers in a diagonal pattern from right to left, you will be able to find the Fibonacci sequence. Below will be a picture of how this is implemented. Another way that this will help in future courses is that it allows you to find squares of a number easily. If you look at the 3rd diagonal row, adding two consecutive numbers from left to right will give the square of a number. A picture of this will also be posted below. Another way that this is implemented in future courses is statistics and probability. This triangle can be used to find the probability of many different things. This is only a few ways that the triangle can be used in future courses, considering that there are plenty of other ways it can be used. In all, this is a very important topic for someone that is pursuing mathematics.

Fibonacci sequence:

Squares of a number:

 

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This video would be a great way to either start a lesson on Pascal’s Triangle or to review the lesson before a test. The video shows different ways that you can implement the triangle to solve different things in mathematics. If this was the video to start the lesson, I would have each student take out a notebook and writing utensil while watching the video. Throughout the video the students would have to find at least three different ways a person may use Pascal’s triangle that they found particularly interesting. This should lead to most of the ways to be picked by at least one student. After they share their answers, explain further why these work. This could make students more intrigued with the subject. If the video was for a review of the topic, I would also have the students have out a writing utensil and a notebook. For this instance, I would have each individual write down what they had forgotten about Pascal’s triangle. From here the teacher will review the points that were most forgotten, serving as a review.

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?”

 

My Favorite One-Liners: Part 32

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. Today’s story is a continuation of yesterday’s post. I call today’s one-liner “Method #1… Method #2.”

Every once in a while, I want my students to figure out that there’s a clever way to do a problem that will save them a lot of time, and they need to think of it.

For example, in Algebra II, Precalculus, or Probability, I might introduce the binomial coefficients to my students, show them the formula for computing them and how they’re related to combinatorics and to Pascal’s triangle, and then ask them to compute \displaystyle {100 \choose 3}. We write down

\displaystyle {100 \choose 3} = \displaystyle \frac{100!}{3!(100-3)!} = \displaystyle \frac{100!}{3! \times 97!}

So this fraction needs to be simplified. So I’ll dramatically announce:

Method #1: Multiply out the top and the bottom.

This produces the desired groans from my students. If possible, then I list other available but undesirable ways of solving the problem.

Method #2: Figure out the 100th row of Pascal’s triangle.

Method #3: List out all of the ways of getting 3 successes in 100 trials.

All of this gets the point across: there’s got to be an easier way to do this. So, finally, I’ll get to what I really want my students to do:

Method #4: Write 100! = 100 \times 99 \times 98 \times  97!, and cancel.

The point of this bit of showman’s patter is to get my students to think about what they should do next as opposed to blindly embarking in a laborious calculation.

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As another example, consider the following problem from Algebra II/Precalculus: “Show that x-1 is a factor of f(x)=x^{78} - 4 x^{37} + 2 x^{15} + 1.”

As I’m writing down the problem on the board, someone will usually call out nervously, “Are you sure you mean x^{78}?” Yes, I’m sure.

“So,” I announce, “how are we going to solve the problem?”

Method #1: Use synthetic division.

Then I’ll make a point of what it would take to write down the procedure of synthetic division for this polynomial of degree 78.

Method #2: (As my students anticipate the real way of doing the problem) Use long division.

Understanding laughter ensures. Eventually, I tell my students — or, sometimes, my students will tell me:

Method #3: Calculate f(1).

 

Lessons from teaching gifted elementary school students: Index (updated)

I’m doing 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 various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

Part 6a: Calculating $(255/256)^x$.

Part 6b: Solving $(255/256)^x = 1/2$ without a calculator.

Part 7a: Estimating the size of a 1000-pound hailstone.

Part 7b: Estimating the size a 1000-pound hailstone.

Part 8a: Statement of an usually triangle summing problem.

Part 8b: Solution using binomial coefficients.

Part 8c: Rearranging the series.

Part 8d: Reindexing to further rearrange the series.

Part 8e: Rewriting using binomial coefficients again.

Part 8f: Finally obtaining the numerical answer.

Part 8g: Extracting the square root of the answer by hand.

Combinatorics and Jason’s Deli: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on an advertisement that I saw in Jason’s Deli.

Part 1: The advertisement for the Jason’s Deli salad bar.

Part 2: Correct calculation of the number of salad bar combinations.

Part 3: Incorrect calculation of how long it would take to eat this many combinations.

 

 

Combinatorics and Jason’s Deli (Part 2)

Jason’s Deli is one of my family’s favorite places for an inexpensive meal. Recently, I saw the following placard at our table advertising their salad bar:

fb_img_1470352586255.jpg

The small print says “Math performed by actual rocket scientist”; let’s see how the rocket scientist actually did this calculation.

The advertisement says that there are 50+ possible ingredients; however, to actually get a single number of combinations, let’s say there are exactly 50 ingredients. Lettuce will serve as the base, and so the 5 ingredients that go on top of the lettuce will need to be chosen from the other 49 ingredients.

Also, order is not important for this problem… for example, it doesn’t matter if the tomatoes go on first or last if tomatoes are selected for the salad.

Therefore, the number of possible ingredients is

\displaystyle {49 \choose 5},

or the number in the 5th column of the 49th row of Pascal’s triangle. Rather than actually finding the 49th row of Pascal’s triangle by direct addition, it’s simpler to use factorials:

\displaystyle {49 \choose 5} = \displaystyle \frac{49!}{5! \times 44!} = \displaystyle \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44!}{5 \times 4 \times 3 \times 2 \times 1 \times 44!}

= \displaystyle \frac{49 \times 48 \times 47 \times 46 \times 45}{5 \times 4 \times 3 \times 2 \times 1}

= 49 \times 12 \times 47 \times 23 \times 3

= 1,906,884.

Under the assumption that there are exactly 50 ingredients, the rocket scientist actually got this right.

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.

PascalProblem

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):

squareroot

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

Lessons from teaching gifted elementary students (Part 8f)

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.

PascalProblem

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)

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.

PascalProblem

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.