In Exercises 11–24, use mathematical induction to prove that each...

Question

In Exercises 11–24, use mathematical induction to prove that each statement is true for every positive integer n.

1 · 2 + 2 · 3 + 3 · 4 + ... + n(n + 1) = n(n + 1)(n + 2)/3

Accepted Answer

Hello. Today we're going to be using mathematical induction to show that the given statement is true. Now the first step in mathematical induction is to show that the right hand side of the statement can at least equal to the first portion or the first term of the summation. On the left hand side. And we can show that by allowing end to equal to one. So letting N equal to one is going to give us the statement one times four is equal to end, which is one times N plus one, which is one plus one times N plus five, which is one plus five. All of that divided by three. Now on the left hand side four times one is just going to give us four. And on the numerator on the right hand side, one plus one is going to give us two and five plus one is going to give us six. So six times two times one is going to give us 12 and that's going to be over three. Finally 12 divides by 34 times. So we have four is equal to four which is a true statement. So the first step shows that the right hand side statement is at least equal to the first term in the left hand side summation. But now we want to come up with the statement that shows that this right hand side is equal to every term in the left hand side. And we can do that by letting end equal to K, which is going to be the second step of the mathematical induction. So letting an equal to K is going to give us the statement one times four plus two times five plus three times eight plus all the terms going to the case term now, which is K times K plus three is equal to K times K plus one Times K-plus five. All that divided by three. So this is a statement that shows that the right hand side is equal to every term in the left hand side. So we're going to assume that this statement is true for now. Now the final step is to show that this statement is true if we add the next term and the next term is going to be when N is equal to K plus one. So adding the next term into our statement is going to give us one times four plus two times five Plus three times plus all the terms going to K times K plus three plus the next term where we have K plus one. And that's going to be the quantity K plus one times K plus one plus three, which is going to be K plus four. And that should equal to K plus one times K plus one plus one, which is K plus two times K plus one plus five, which is K plus six. All of that divided by three. Now the final step of the mathematical induction is to show that the left hand side is equal to the right hand side. But how can we do that if we don't even know how many terms are given to us? Well, the tricky part about this is to notice that this part of the left hand side is equivalent to the previous statement that we came up with, with that being said, we can use the previous statement as a substitution for this part of the left hand side, and we can let this part equal to K times K plus one times K plus 5/3. So doing this is going to give us K times K plus one times K plus 5/ Plus K-plus one Times K-plus four. And that's going to equal to K plus one times K plus two times K plus six. All over three. Now we want to show that the left hand side is equal to the right hand side. And in order to do that, we're just going to use straight straight algebra. And evaluating the left hand side. Well, on the left hand side of the statement, we have K times K plus one Times K-plus five over three Plus K-plus one times K plus four. And for now we're going to put these quantities over one. So we want to go ahead and add these quantities together and in order to do that, we need to have two fractions of the same denominator here, the lowest common denominator is going to be three. And in order to get three on the right hand side of the denominator, we're gonna multiply this fraction by 3/3. And doing this is going to give us K times K plus one times K plus five over three plus three times K plus one Times K-plus four. All of that over three. Now that we have the same denominator in both the fractions we can go ahead and combine the fractions together into a single fraction and that's going to give us K times K plus one times K plus five plus three times K plus one times K plus four. All of that over three. Now notice that we have the factors of K plus one in both of the terms here. So we can go ahead and factor out a K plus one from the entirety of this quantity. And doing so is going to give us K plus one times K times K-plus five plus three times K plus four, All of that over three. Now, in order to simplify the numerator further, we're going to distribute K into K plus five and distribute three into K plus four. This is going to give us K plus one times the quantity, K squared plus five, K plus three K plus 12. All of that over three and combining like terms we can go ahead and combine five K with three K. And that's going to give us K squared plus eight K plus 12. Now this try no meal here is factory able as it factors into the quantities K plus two and K plus six. So factoring K squared plus eight K plus 12 is going to give us K plus one Times K-plus two Times K-plus six. All of that divided by three. And that's going to be the simplified version of the left hand side of the statement. Now notice that the left hand side is equal to the right hand side. That shows that the statement is true for every integer end in the original statement and with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to perform mathematical induction and I'll go ahead and see you all in the next video.

In Exercises 11–24, use mathematical induction to prove that each statement is true for every positive integer n. 1 · 2 + 2 · 3 + 3 · 4 + ... + n(n + 1) = n(n + 1)(n + 2)/3

Key Concepts

Mathematical Induction

Summation of Series

Polynomial Functions

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