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.

4 + 8 + 12 + ... + 4n = 2n(n + 1)

Accepted Answer

hey everyone here we are asked to prove that the given statement is true for every positive integer N. Using mathematical induction. So our given statement is five plus 10 plus 15 plus some other values plus five N is equal to five halves times N times the quantity of n plus one. So our first step here in this problem is to prove that this given statement is true for when N is equal to one. So doing this, we need to prove that the left hand side is equal to the right hand side, so beginning with our left hand side, since N is equal to one, we need to select the first term in our sequence here. So we have five on the left hand side and for the right hand side we need to use this expression on the right hand side from our given statement and replace the N variables with one. So we'll have five is equal to five halves times one times the quantity of one plus one. And now simplifying, we see that five is equal to five and that the left hand side is in fact equal to the right hand side. And therefore the given statement is indeed true for when N is equal to one. And so now we can move on to the second step and we need to let this given statement be true for when N is equal to K. So now we just need to substitute in K for four N in our given statement. So we have five plus 10 plus 15 plus five K is equal to five halves K times the quantity of K plus one. And now from here we need to prove that the given statement is true for when N is equal to K plus one. And so now we need to rewrite the given statement substituting in this K plus one. So we have five plus 10 plus plus five K. We keep this five K from before but we need to add the name next term. This K plus one term. So we add five times the quantity of K plus one is equal to five halves times the quantity of K plus one times the quantity of K plus one plus one. And now from here we again need to work to get the left hand side equal to the right hand side. Now looking at our current statement so far, we see that we have this sequence five plus 10 plus 15 plus five K. Now this is the same sequence from our above statement where we sub in K for N. And we see that this sequence is equal to five halves K times the quantity of K plus one. So we can take this expression and substituted into our current statement. So now rewriting, we have five halves times K times the quantity of K plus one plus five times the quantity of K plus one is equal to five halves times the quantity of K plus one. And now simplifying this last set of parentheses, we have times the quantity of K plus two. So now we need to continue to manipulate the left hand side. We see that we have K plus one for both parts of the expression here. So we can factor it out. So we have the quantity of K plus one times the quantity of five halves, K plus five is equal to the right hand side here which is five halves times the quantity of K plus one times the quantity of K plus two. Now further factoring on our left hand side, further manipulation, we can factor out this five halves and this gives us five halves times the quantity of K plus one. And now we have times the quantity of K plus two and then again for the right hand side we now have the same expression here. We have five halves times the quantity of K plus one, times the quantity of K plus two. And with this we can clearly see that the left hand side is indeed equal to the right hand side. And with this we know that the given statement is in fact true for when N is equal to K plus one. And because of this we now know that the given statement is true for every positive integer N. Therefore we are left with answer choice D. Thanks for watching. I hope you found this video helpful and I'll see you again next time

In Exercises 11–24, use mathematical induction to prove that each statement is true for every positive integer n. 4 + 8 + 12 + ... + 4n = 2n(n + 1)

Key Concepts

Mathematical Induction

Summation of an Arithmetic Series

Quadratic Functions

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