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Algebra II Textbook26 chapters | 256 lessons

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Lesson Transcript

Instructor:
*Yuanxin (Amy) Yang Alcocer*

Amy has a master's degree in secondary education and has taught math at a public charter high school.

Watch this video lesson to learn about mathematical induction and how you can use it to prove mathematical statements. See how it is similar to falling dominoes.

In this video lesson, we talk about mathematical induction. What exactly is it? **Mathematical induction** is a way of proving a mathematical statement by saying that if the first case is true, then all other cases are true, too. So, think of a chain of dominoes. If you tip the first domino, what happens to all the other dominoes? They fall, too. And there we have an example of mathematical induction in real life. If the first domino falls, then all the other dominoes fall, too.

Mathematical induction has two steps to it. The first is to prove that our first case is true. The second is to prove that if any other case is true, then the following case is also true. It's like a chain effect. If any one case is true, then the next is true also. And if this is the case, then it means that all the cases in any one particular problem are true. Just like with our falling dominoes, if the first domino falls, then all the dominoes will fall because if any one domino falls, it means that the next domino will fall, too.

So, how do we use mathematical induction? We use it to prove five mathematical statements, such as 1 + 2 + 3 + 4 + . . . + *n* = (*n*)(*n* + 1) / 2 is true for all *n*. There are two steps to using mathematical induction.

- Show the first case, usually
*n*= 1, is true. - Assume that the case
*n*=*k*is true, so therefore the case*n*=*k*+ 1 is also true.

So, let's see how we go about using mathematical induction. Why don't we go ahead and try to prove the statement 1 + 2 + 3 + 4 + . . . + *n* = (*n*)(*n* + 1) / 2?

- We start by showing that the case
*n*= 1 is true. When*n*= 1, our statement becomes 1 = (1)(1 + 1) / 2. Evaluating this, we get 1 = (1)(2) / 2, which equals 1 = 2 / 2. This then becomes 1 = 1. Is that a true statement? Yes, it is, and we have proved our first case. - The second step is kind of tricky. We are assuming that the case
*n*=*k*is true. So, we have the statement 1 + 2 + 3 + 4 + . . . +*k*= (*k*)(*k*+ 1) / 2 is true. Now we need to show that if this case is true, then so is the case*n*=*k*+ 1. The case*n*=*k*+ 1 changes the statement to 1 + 2 + 3 + 4 . . . +*k*+ (*k*+ 1) = (*k*+ 1)((*k*+ 1) + 1) / 2.

To prove that this statement is true, we can use our assumption that the case *n* = *k* is true. Notice that the terms all the way back up to the *k* + 1 term make up the *n* = *k* case, so we can replace all those terms with what they equal, which is (*k*)(*k* + 1) / 2. So, now the statement that we need to prove becomes (*k*)(*k* + 1) / 2 + (*k* + 1) = (*k*+1)((*k* + 1) + 1) / 2. Let's add and multiply everything out on both sides and see if they will equal each other. If they equal each other, then we will have proved our statement is true.

Are both sides equal to each other? Yes, they are! We proved that our mathematical statement 1 + 2 + 3 + 4 + . . . + *n* = (*n*)(*n* + 1) / 2 is true. Do you see the dominoes falling into place?

Let's look at another problem. Let's prove the statement 1 + 3 + 5 + . . . + (2*n* - 1) = *n*^2.

- We begin by showing that the case
*n*= 1 is true. We have 1 = 1^2, which becomes 1 = 1. Is this true? Yes, both sides are the same. - Next, we assume that the case
*n*=*k*is true. So, we have 1 + 3 + 5 + . . . + (2*k*- 1) =*k*^2 is true. We will use this to show that the case*n*=*k*+ 1 is true. The case*n*=*k*+ 1 is 1 + 3 + 5 + . . . + (2*k*- 1) + (2(*k*+ 1) - 1) = (*k*+ 1)^2. We can see that all the terms before the last term in our series equals the case*n*=*k*. So, we make that replacement since we know that all those terms equal*k*^2. So, our statement becomes*k*^2 + (2(*k*+ 1) - 1) = (*k*+ 1)^2. Now, we add and multiply out both sides to see if they are equal.

Are the two sides equal to each other? Yes! So, that means the statement 1 + 3 + 5 + . . . + (2*n* - 1) = *n*^2 is true. Again, we have falling dominoes at work!

What have we learned? We've learned that **mathematical induction** is a way of proving a mathematical statement by saying that if the first case is true, then all other cases are true, too. Think of falling dominoes. If you tip the first domino, then all the other dominoes will fall. The two steps to using mathematical induction are:

- Show that the first case, usually
*n*= 1, is true. - Assume that the case
*n*=*k*is true, so therefore the case*n*=*k*+ 1 is also true.

The second is best done by using the assumption that the case *n* = *k* is true. Because we can assume this case to be true, we can replace this part with what it equals when we try to prove that the case *n* = *k* + 1 is true.

After this lesson, you'll be able to:

- Define mathematical induction
- List the steps to using mathematical induction
- Prove a statement is true using those steps

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14 in chapter 21 of the course:

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Algebra II Textbook26 chapters | 256 lessons

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- How to Use Factorial Notation: Process and Examples 4:40
- How to Use Series and Summation Notation: Process and Examples 4:16
- Arithmetic Sequences: Definition & Finding the Common Difference 5:55
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- The Sum of the First n Terms of an Arithmetic Sequence 6:00
- Understanding Arithmetic Series in Algebra 6:17
- Working with Geometric Sequences 5:26
- How and Why to Use the General Term of a Geometric Sequence 5:14
- The Sum of the First n Terms of a Geometric Sequence 4:57
- Understand the Formula for Infinite Geometric Series 4:41
- Using Recursive Rules for Arithmetic, Algebraic & Geometric Sequences 5:52
- Using Sigma Notation for the Sum of a Series 4:44
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