Use the Principle of Mathematical Induction to show that the given statement is true for all nat

Use the Principle of Mathematical Induction to show that the...

Key Concepts

Inductive Step

This is the second key step where one shows that if the statement holds for a natural number n (as assumed in the inductive hypothesis), then it necessarily holds for the next number, n+1. This step bridges the base case and all subsequent cases.

Inductive Hypothesis

In the induction proof, this is the assumption that the statement holds for an arbitrary natural number n. This hypothesis is then used as a stepping stone to derive the truth of the statement for n+1.

Algebraic Manipulation

This involves using algebraic techniques to simplify, re-arrange, and combine expressions in the process of the inductive step. It is crucial for proving that the assumed formula extends correctly from n to n+1.

Principle of Mathematical Induction

This is a proof technique used to establish that a statement is true for all natural numbers. It involves showing that the statement holds for an initial case (usually the number 1) and then demonstrating that if the statement holds for an arbitrary natural number k, it must also hold for k+1.

Base Case

This is the first step in the induction process where the statement is verified for the initial value, typically n = 1. Establishing the base case confirms that the property holds at the starting point of the induction.

Transcript

00:01 Okay, so we want to use mathematical induction to show that the following is true.

00:04 So let's start with condition 1.

00:06 We need to show that n is equal to 1 is true.

00:09 So starting with our left hand side, we have 2 times 1, minus 1, and then we have 2 times 1.

00:15 So that's equal to a 1 times 2, which is equal to 2.

00:19 And is that equal to our right hand side? well, let's see.

00:25 So that's equal to a 3, and this is a 2.

00:29 So we get 6 over 3, which is equal to 2.

00:32 So we see that this is true.

00:34 And now let's move on to condition 2.

00:37 So in this case, we need to assume that n is equal to k is true.

00:43 So let's write out n is equal to k.

00:45 So we have 1 times 2 and then 3 times 4, so on, up to 2k minus 1 times a 2k.

00:55 And that's going to be equal to what we have up there.

00:58 So instead of an n, we have a k.

01:05 Okay.

01:05 And now we want to show that n is equal to k plus 1 is true.

01:09 So let's write out how it would luck if it was true.

01:12 We'd have 1 times 2 plus 3 times 4 all the way up to.

01:17 I'm going to include our k term.

01:20 So we have that should be a 2k minus 1 times a 2k and then plus our k plus 1.

01:29 So this is a 2k plus 2 minus 1.

01:31 So we get 2k plus 1.

01:33 And then we have a 2k plus 2.

01:38 So that's going to be equal to 1 over 3, k plus 1, k plus 2, and then 4k plus 4 minus, so we get 4k minus, or sorry, that's a plus 3.

01:50 Okay, so starting with our left hand side, let's notice that this entire portion is n is equal to k, and we assume that n is equal to k is true, so we can rewrite it like so...

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