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.3 + 7 + 11 + ... + (4n - 1) = n(2n + 1)

Accepted Answer

Hello. Today we're going to be showing that the given statement is true through mathematical induction. So what we are given is five plus 12 plus 19 plus all the terms to the end term seven and minus two. And this should be equal to the statement seven and squared plus three N over two. Now, how do we show that this is true? While in mathematical induction, the first step is to show that the given statement here is at least equal to the first term in the summation. And in order to do that, we're going to allow end to equal to one. So doing this is going to give us the statement five is equal to seven and squared, which is going to be seven times one squared plus three. End, which is three times one. All of that divided by two. So let's just go ahead and simplify the right hand side for now one race to any power is just going to give us one and seven times one is just going to be seven and three times one is just going to give us three. So the right hand side simplifies to seven plus three divided by two and seven plus three is going to give us 10. So we have five is equal to 10/2, which simplifies to five. So this statement shows that the original statement given to us is at least equal to the first term in the summation. Now, the second step in mathematical induction is to create a statement that shows that at least the right hand side is equal to every term in the summation. And in order to do that, the second step states that we're going to let N equal to K. When N is equal decay, we get the statement five plus 12 plus plus all of the terms to the end term while not anymore, but K term seven K minus two. And that is going to equal to seven K squared plus three K. All of that divided by two. Now, this is assuming that the right hand side is equal to every term in the summation. So we're going to assume that this statement is true for now. Now, the third step of mathematical induction is to show that the statement that we came up with in the second step is going to equal to the next term in the summation. So we're going to be adding another term in the given summation. And that term is going to be when n is equal to K plus one. So adding the K plus one term is going to give us five plus plus 19 plus all of the terms up until seven k minus two. And now we add our K plus one term, which is going to be seven times k minus K plus one minus two. And that is going to equal to the right hand side. But with now K plus one, which is going to be seven, seven times K plus one squared plus three times K plus one. all of that over to. So now we need to show that the left hand side is equal to the right hand side. But how can we show that if we don't even know how many terms are in the left hand side? Well, here's the tricky thing about the mathematical induction. Notice that this part of the left hand side of the statement is equivalent to the statement that we came up with in the second step. With that being said, we can use the second step statement as a substitution for this portion of the left hand side, so we can substitute this portion with seven K squared plus three K plus two. And making that substitution is going to give us the statement seven K squared plus three K over two plus while we're going to distribute seven into K plus one, so that's gonna be plus seven K plus seven minus two. And that should equal to seven times K plus one squared plus three times K plus 1/2. Now, at this point it's just going to be algebra to show that the left hand side is equal to the right hand side. Now there is a little bit of simplification that we need to do here. So let's go ahead and start with the right hand side. The right hand side, we have seven times K plus one squared plus three times K plus one, all over two. So in order to simplify the right hand side, we're going to need to foil out K plus one squared and we're going to need to distribute +32 K plus one. Doing this is going to give us seven times K squared plus two K plus one plus three K plus three. All of that. Over to the next step. In the simplification is to distribute seven into K squared plus two K plus one. And that's going to give us seven K squared plus 14 K plus seven plus three K plus three. All that over to next. We want to go ahead and combine the like terms which we can combine 14 K with three K and seven with three. And the right hand side is going to simplify to seven K squared plus 17 K plus 10. All of that over to. Now let's go ahead and simplify the left hand side. On the left hand side. We have seven k squared plus three K over two plus seven K plus seven minus two, combining seven. And negative two is going to give us positive five. So we have the quantity seven k squared plus three K over two plus the quantity seven k plus five. Now seven K plus five can be put over one. And in order to combine these fractions together we can multiply this quantity by 2/2. This is going to give us seven k squared plus three K over two plus two times the quantity seven k plus 5/2. Now we can go ahead and combine these fractions together and doing this is going to give us seven K squared plus three K plus two times the quantity seven K plus five, all of that over to. And in order to simplify the numerator here, we're going to distribute two into seven K plus five. That's going to give us seven k squared plus three K plus 14 K plus over to. And now we just need to go ahead and combine like terms, we can go ahead and combine 14 K with three K. And that's going to give us seven K squared plus 17 K plus 10/2. And that's going to be the simplified version of the left hand side. Now we just want to go ahead and set these equal to each other. So the right hand side is going to be seven k squared plus 17 K plus 10/2. And that is equal to the left hand side, which is seven k squared plus 17 K plus 10/2. Since both of the sides are equivalent to each other, this means that their statement is true and that means that the original statement is true for all positive integers N. 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.3 + 7 + 11 + ... + (4n - 1) = n(2n + 1)

Key Concepts

Mathematical Induction

Arithmetic Series

Algebraic Manipulation

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