Difficult Math Need not be Difficult

Do you avoid math because you think it is too difficult? Here's an easy derivation of a powerful Pre-calculus formula that just might change your mind.

On the first day of school I would always share a secret with my math students. It would go like this:

Let me explain something to you about mathematicians. They are inherently lazy. When presented with a task the first thing they always look for is a shortcut. Their primary goal in life is to do the least amount of work possible to determine whatever it is they are required to find out. Here’s the bottom line—the less time they take to solve problems the more time they have to play video games or whatever it was that Pythagoras and Pascal did in their spare time.

I would pause momentarily. 

For the life of me I don’t understand why you teenagers don’t just love math—it’s so you. It represents so many of the traits that high school students value.

This opening always resulted in nervous laughter throughout the room. It was now time for me to close the deal.

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Overcoming math anxiety

The culture of this country is far from positive when it comes to utilizing mathematics. During these first few minutes of the class I needed to establish that the curriculum of the upcoming year had practical applications while not being ridiculously difficult or complicated. In addition, the validity of the label “lazy” for people who sang the praises of math had to be demonstrated.

Turning the complex into simple

It begins with a statement I have used to terrorize parents at Back to School Night, non-math colleagues at workshops and my students on that first day. 

Let’s start this lesson by deriving a formula that is critical to the study of Pre-calculus. What better place to begin than finding the specific term of an arithmetic sequence of numbers?

I would then write a sample problem on the board:What is the356th term of the arithmetic sequence 21, 27, 33, 39 …?

I would then explain that there are two available options for finding the solution. 

One way we could solve this would be to get two very sharp pencils and a really large sheet of paper and, well, you know the rest. Or you could think like a mathematician and find some easy shortcut.

Starting with the basics

A sequence is defined to be a set of numbers that have a discernible pattern which can be used to predict subsequent numbers. The following sets would be examples of sequences (the … indicates that the pattern continues indefinitely):

1, 7, 13, 19, 25 … (The addition of six is the pattern and the next number would be 31.)

3, 6, 12, 24, 48 … (The multiplication by two is the pattern and the next number would be 96.)

1, 2, 4, 7, 11, 16 …. (The addition of one to first number, two to the next, and three to the third, etc. is the pattern. The number after 16 would be 22.)

1, 4, 9, 16, 25 ... (The pattern is one squared, two squared, three squared, etc. The number after 25 would then be 36.)

There are an infinite number of ways that a sequence can be formed. One such way is called an Arithmetic Sequence which is defined to be a set of numbers created by the addition of the same value to each “term”. (Note: term is math-talk for a number.) This constant is called “the common difference” (d) since it can be determined by subtracting any term from the one that follows it. Thus, two typical arithmetic sequences would be:

4, 9, 14, 19 … (d=5)

18, 8, -2, -12 … (d=-10)

All arithmetic sequences consist of a first term (a1), an nth term (an), the number of terms (n) and the number that is added each time called “the common difference” (d).

Creating a mathematical shortcut

In its most basic form creating any arithmetic sequence would look like this:

a1 = a1 + 0d

a2 = a1 + 1d

a3 = a1 + d + d = a1 + 2d

a4 = a1 + d + d + d = a1 + 3d

Simplifying, the sequence would be:

a + 0d, a + 1d, a + 2d, a + 3d …

A pattern is readily apparent. The number preceding “d” (the coefficient) for the first term is zero, for the second it is one, for the third it is two. Every successive term is the first term plus d multiplied by one less than the relative position of the term in the sequence. A formula can now be created:

an (the subscribe "n" indicates specified term) = a1 + (n – 1) x d

Mission accomplished

Now it is time to find the 356th term of the arithmetic sequence 21, 27, 33, 39 …

I will pass on writing out 356 consecutive numbers on that very large sheet of paper and opt for the formula. The first term is 21, the common difference is 6 and the number of terms is 356.

an = 21 + (356 – 1) x 6 = 21 + (355 x 6) = 21 + 21 30 = 2151

About 800 words after we began we have found the correct answer and so much more. A powerful formula from the upper levels of high school math has been developed with a minimum of angst and pain. It is now permanently available to be utilized for solving a variety of values in multiple situations. No pencils need to be repeatedly sharpened. Paper has been conserved. And, of course, significant amounts of time and energy have been saved.