# Mathematical Induction

## Mathematical Induction - How it Works - Video

### Definitions

Definitions:

The principle of mathematical induction needs to have two statements so that it works.

Statement 1 - Pn is true.

Statement 2 - Whenever k is a positive integer such that Pk is true, then Pk+1 is also true.

Steps in Applying:

The principle of mathematical induction needs to have two statements so that it works.

Show - Statement 1 - Pn is true.

Assume - Statement 2 - Whenever k is a positive integer such that Pk is true, then Prove Pk+1 is also true.

### Example 1 - Explore

Example 1 Explore:

If are not sure what to first, then try plugging in some numbers to see if you can see a pattern. After substituting 1, 2, and 3 we found that the statement 1 + 2 + 3 + ... + n = [n * (n + 1)] / 2.

The 1st term

=> 3 *(-1/4)1-1

=> 3 *(-1/4)0

=> 3 * 1

=> 3

The 2nd term

=> 3 *(-1/4)2-1

=> 3 *(-1/4)1

=> 3 * (-1/4)

=> (-3/4)

The 3rd term

=> 3 *(-1/4)3-1

=> 3 *(-1/4)2

=> 3 * (1/16)

=> 3/16

The 10th term

=> 3 *(-1/4)10-1

=> 3 *(-1/4)9

=> 3 * (-1/262,144)

=> -3/262,144

### Example 1 Step 1 & Step 2

Example 2:

Luckily for us, when we explored, we already did the first step and prove that P1 is true. Now we substitute k into our statement and assume that is true. Finally we want to prove k + 1 is true so we down our goal.

### Example 1 Step 2:

Example 1 Step 2:

Here we have all the steps to get our assumption to our goal.

1st => We put parentheses around the first times before k + 1

2nd => We substituted the induction hypothesis

3rd => We found a common denominator.

4th => We group the fractions together.

5th => We factored the (k + 1) term since it is on both sides of the addition sign.

6th => We changed (k + 2) to match our goal.

### Example 1 Final Part:

Example 1 Final Part:

We have shown that P1 is true and we proving Pk+1 is true.

Therefore we can say that Pk+1 is true by mathematical induction.