Write out in full and verify the statements S1, S2, S3, S4, and S5 for the following. Then use m

Write out in full and verify the statements S1, S2, S3, S4,...

Key Concepts

Verification of Initial Cases

Before applying the induction process, it is common practice to verify the statement for several small positive integers (e.g., S1, S2, S3, S4, and S5). Although not a formal part of the induction proof, this verification provides insight and lends credence to the general truth of the statement and can help identify potential mistakes early in the process.

Inductive Step

The inductive step is the part of the proof where, under the assumption that the statement is true for n = k (the inductive hypothesis), one shows that the statement must also be true for n = k + 1. Successfully proving this step completes the induction argument, confirming that the statement holds for all positive integers.

Inductive Hypothesis

The inductive hypothesis is the assumption that the statement is true for some arbitrary positive integer k. This assumption is used as a stepping stone in the proof, allowing us to demonstrate that if the statement holds for n = k, then it must also hold for n = k + 1.

Base Case

The base case involves verifying that the statement holds for the first element in the series of integers, typically n = 1. This step is crucial because it establishes the truth of the statement at the starting point, which is necessary before induction can be applied to reach further conclusions.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a statement holds for all positive integers. It involves two main steps: first, proving the statement for an initial case (often n = 1), and second, assuming the statement is true for some arbitrary positive integer k (the inductive hypothesis) and then proving it for the next integer, k + 1 (the inductive step). This method effectively 'propagates' the truth of the statement from one integer to the next.

Transcript

00:02 And in this problem we are going to write out and verify the statements s1, s2, s3, s4 and s5.

00:13 Then we use mathematical induction to prove that the statement is true for every positive integer n.

00:22 In this case, the statement sn is 1 plus 3 plus 5 up to 2n minus 1 is equal to n square.

00:34 So let's see for a moment this statement in detail.

00:40 And as we can see here, each of these numbers are odd numbers.

00:49 And each one can be written just this way for successive value of n.

00:56 So the first one, this term here, 1, is 2 times 1 minus 1.

01:03 So this expression here is equivalent to this number 1.

01:06 If we do the calculations, we can verify that.

01:10 2 times 1 is 2, minus 1 is 1 plus.

01:14 Then 3 is this same formula, you see now the factor 2 here.

01:21 2 times 2 plus 1, sorry.

01:26 And so, and that is true because we have 2 times 2 is 4 minus 1 is 3.

01:40 The next 1 is 2 times 3, which is the next positive 0.

01:47 Into the n minus 1.

01:51 Now this is 6 minus 1, it's 5.

01:55 So these numbers are odd numbers and each one can be written this way for the successive values of n.

02:09 And now if we count the terms, we can see that this factor, which multiplies the number 2, is the counting factor, that is this is the first term, this is the second term, this is the third term.

02:29 And so this is going to be the nth term.

02:35 This is the first, this is the second, this is the third, that is the nth term.

02:42 That is we have n terms, this is in this sum.

02:55 N terms.

02:57 And that's an important fact because we can say then that the statement as n is equivalent to saying that the sum of the first n -odd positive numbers is equal to that quantity n, that quantity of numbers, n squared.

03:24 And so, s1, for example, we are going to write s -1, in this case will be only one term.

03:35 And this will be the first term of the sum that is 1 here.

03:41 So in this case we have 1 as a result of the sum because the sum consists of only the term 1 and the right side will be 1 square.

03:51 That is the number of terms that is 1 square.

03:57 And this is 1 equal 1 which is true.

04:02 So we have verified the statement for n equals 1.

04:06 For n equals 2, we have now two terms in the sum, that is 1, which is the first odd number, plus 3, which is the second odd number.

04:18 And on the right side, we have the number of terms that is n is 2, and that can be squared.

04:26 And then we have 4, which is 1 plus 3, and 2 squared is 4 also.

04:32 So it's again a true statement.

04:34 If we write as 3, we have the sum of the first three odd numbers, that is 1 plus 3 plus 5.

04:45 On the left and on the right we have n, that is the number of terms 3 square.

04:54 And then here we have 8 plus 1, 9, and 3 squared is 9 also.

05:00 So it's again a true statement.

05:02 Is 4 will be the sum of the first.

05:08 4 odd numbers 1 plus 3 plus 5 plus 7.

05:13 And on the right side we have n squared, that is 4 square.

05:23 And if we sum up these numbers, we have 12, 7 plus 5, 12 plus 15 plus 1 is 16.

05:32 And 4 squared is 4 times 4 is 16.

05:36 So it's true...

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