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

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 statement on the right hand side is at least equal to the first term on the left hand side. And we can do this by showing that N is equal to one. So you letting N equal to one is going to give us the statement 1/1 times three is equal to one times three times one plus 5/4 times one plus one times one plus two. Now on the left hand side in the denominator one times three is just going to give us three. So we have one third on the left hand side and in the numerator on the right hand side three times one is just going to give us 33 plus five is eight and times one is still going to be eight, we have eight in the numerator. And for the denominator one plus one is going to give us two and one plus two is going to give us three. So we have four times two times three, which is going to give us 24. And finally 8/24 can simplify to one third. So we have this statement 1/3 is equal to 1/3, which is true. So this at least shows that the statement on the right hand side is at least equal to the first term on the left hand side. Now the second step in mathematical induction is to make a statement that shows that the right hand side is at least equal to every term on the left hand side. And we can do that by letting n equal to K when N is equal to K. We get the statement 1/1 times three plus 1/2 times four plus 1/ times five plus all of the terms going to the case term now, which is one over K times K plus two. And that is going to equal to K times three. K plus 5/4 times K plus one Times K-plus two. Now, this is a statement that shows that the right hand side is at least equal to every term on the left hand side. So we're going to assume that this statement is true for now. Finally, the third step of the mathematical induction is to show that this statement is true. If we add the next term of the summation and the next term is where N is equal to K plus one. So adding this next term is going to give us the statement 1/1 times three plus 1/2 times four plus 1/ times five plus all the terms going to the case term, which is one over K times K plus two plus the next term, where we have K plus one. And that's going to be one over K plus one times the quantity K plus one plus two, which is going to be K plus three. And that is going to equal to the quantity K plus one times three times the quantity K plus one plus five over four times K plus one plus one, which is going to be K plus two times K plus one plus two, which is going to be K plus three. Now we want to show that the left hand side of the statement is equal to the right hand side of the statement. But how can we do that if we don't even know how many terms we have in the left hand side? Well, notice one thing in the mathematical induction, this portion of the left hand side statement is equivalent to the statement we came up with in the second step. That means we can use the statement. And the second step as a substitution for this portion of the left hand side. And we're going to substitute this portion with K times three. K plus 5/4 times K plus one times K plus two. So making the substitution is going to give us the statement K times three. K plus 5/4 times K plus one times K plus two plus the quantity one over K plus one Times K-plus three. And this should equal to the right hand side which is going to be K plus one. And let's go ahead and distribute three into K plus one. That's gonna be K plus one times the quantity three K plus three plus five and three plus five is going to give us eight. So we have three K plus eight. And that's over four times K plus two Times K-plus three. Since the right hand side is in its simplest form, we want to go ahead and show that the left hand side is equal to the right hand side. And at this point we can just use our algebra to show that. So let's go ahead and simplify the left hand side with a statement. Once again, the left hand side is K times three K plus 5/4 times K plus one Times K-plus two plus one over K plus one Times K-plus three. We want to go ahead and combine these fractions together. So we're gonna multiply each fraction by a quantity. That way we can be able to get the lowest common denominator in both of the quantities. Now, the L. C. D. Is the all the possible combination of the product of any of the given terms in the denominators. So the L C. D is going to be four times K plus one times K plus two times K plus three. So we need to multiply each fraction by the by the quantity that it's missing on the right hand side. This fraction is missing four times K plus two in the denominator. So we're going to multiply this fraction by four times K plus two. Over four times K plus two. And on the left hand side the denominator is missing K plus three. So we're gonna multiply the numerator by K plus three and the denominator by K plus three. And doing this is going to give us K times three K plus five times K plus three over four times K plus one times K plus two times K plus three. And we're going to add four times K plus two over four times K plus one times K plus two times K plus three. Alright, now that we have the same denominator in both of the fractions, we can go ahead and combine the fractions into a single fraction and that's going to give us K times three K plus five times K plus three plus four times K plus two. All of that over four times K plus one times K plus two times K plus three. Now we need to go ahead and simplify the numerator in order to simplify the numerator, we're going to go ahead and foil together the three quantities K times three K plus five plus times K plus three. And we're going to distribute four into K plus two. Doing this is going to give us three k squared plus 14 K squared. I'm sorry, three K cubed plus 14 K squared plus 15 K plus four K plus eight. And all that is going to be over four times K plus one times K plus two times K plus three. Now the numerator is factory Ble. And the numerator is going to factor into the quantities K plus one squared times three K plus eight. And once again, that's over our denominator which is four times K plus one times K plus two times K plus three. Now notice that we have two instances of K plus one in the numerator and we have one instance of it in the denominator. With that being said, we can cancel one of the quantities of K plus one in both the numerator and denominator. And that's going to leave us with K plus one times three. K plus eight. Over four times K plus two Times K-plus three. And notice that the left hand side is equivalent to the right hand side of the statement. So this shows that the original statement is true through mathematical induction. And with that being said, the answer to this problem is going to be D. So I hope this video helps you understand how to do 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/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ... + 1/(n(n+1)) = n/(n + 1)

Key Concepts

Mathematical Induction

Series and Summation

Partial Fraction Decomposition

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