Use mathematical induction to prove that each statement is tru...

Question

Use mathematical induction to prove that each statement is true for every positive integer n. $\sum_{i = 1}^{n} 5 \cdot 6^{i} = 6 (6^{n} - 1)$

Accepted Answer

Hello. Today we're going to prove that the given statement is true. Using mathematical induction. So the first step in the mathematical induction is to show that the right hand side of the statement is equal to the left hand side. Specifically the first term the left hand side. We have the summation beginning from one and going to end of three times four raised to the power of I. So the first term of the summation here is going to be when I is equal to one, so when I is equal to one, we get three times four raised to the power of one and that's going to equal to the right hand side when n is equal to one And when N is equal to one, we get four times four raised to the power of 1 -1. Now all we need to do is simplify the statement four to the power of one is just going to be four. So we have four times three on the left hand side, which is going to give us 12 on the right hand side. Once again, four to the power of one is just going to give us four and four minus one gives us three. So we have four times three and four times three gives us 12. So we have the statement 12 is equal to 12, which is a true statement. So this at least shows that the right hand side is at least equal to the first term of the left hand side summation. Now the second step is to create a statement that shows that the right hand side can be equal to any term in the summation here. And we're going to do that by allowing end to equal to K. So when N is equal to K, we get the statement, the summation from I is equal to 12 k of three times four to the power of I is equal to four times four to the power of K -1. Now, since we're saying that the right hand side is equal to any term in the left hand side, we're going to assume that this statement is true. And the next part of the mathematical induction is to show that the right hand side is true when we add the next term in the summation. And that means that we're going to allow n two equal to K plus one for the next term in the summation. And that's going to give us the summation from n is equal to + K of three times forward to the power of I plus the summation of I is equal to one to k plus one of three times four to the power of I. And that's going to equal to four times four to the power of K plus one minus one. Now, our job here is to show that the left hand side is equal to the right hand side. And the tricky part about this mathematical induction is notice that the first portion of the left hand side is equal to the statement that we came up with in part two. That means that we can go ahead and substitute this left hand side With four times 4 to the power of K -1. So doing the substitution is going to give us four times forward to the power of K -1. And we want to go ahead and use this summation but only use it for the K term because we only want the last term in the summation And that's going to give us three while we're going to add three times four to the power of K plus one and that is equal to the right hand side, which is four times four to the power of K plus one minus one. Now, since we have a simplified version of the left hand side, we want to show that the left hand side is equal to the right hand side. So let's go ahead and just work out the left hand side by itself. So we have four times four to the power of K - plus three times four to the power of K plus one. Now, in order to simplify this, we need to distribute four into the quantity four to the power of k minus one. And that's going to give us four K or four to the power of K times four minus four plus three times four to the power of K plus one. Now in the first part of the statement we have four to the power of K times +44 has the exponents of one. So we can combine these two terms using our exponential multiplication and when we multiply these terms together, we're just going to add the exponents together. That's going to give us four to the power of K plus one minus four, plus three times four to the power of K plus one. And I'm gonna go ahead and reorder this to allow the terms with four to the power of K plus one to be next to each other. Doing this is going to give us four to the power of K plus one plus three times four to the power of K plus one minus four. Now notice that we have four to the power of K plus one. In the first two terms we can go ahead and factor out four to the power of K plus one. And that's going to give us four to the power of K plus one times one plus three minus four. Now you may be thinking to yourself, where did the one come from? Well that's because in the previous step four to the power of K plus one has the coefficient of one. And when you factor out the four term, you're going to give four to the power of K plus one times the coefficients which is one plus three and one plus three is going to give us four. So we have four times four to the power of K plus one minus four. And notice that we have four in the first term and second term, we can go ahead and factor it out and that's going to give us four times the quantity four to the power of K plus one minus one. And notice that the left hand side matches the right hand side of the statement. This shows that the original statement end is true for every term in the summation, and that means that the answer to this problem is going to be D. So I hope this video helps in understanding how to perform the mathematical induction and I'll go ahead and see you all in the next video.

Use mathematical induction to prove that each statement is true for every positive integer n. $\sum_{i = 1}^{n} 5 \cdot 6^{i} = 6 (6^{n} - 1)$

Key Concepts

Mathematical Induction

Summation of Geometric Series

Algebraic Manipulation

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Use mathematical induction to prove that each statement is true for every positive integer n. ∑i=1n5⋅6i=6(6n−1)\sum_{i=1}^{n} 5 \cdot 6^i = 6(6^n - 1)

Key Concepts

Mathematical Induction

Summation of Geometric Series

Algebraic Manipulation

Watch next

Use mathematical induction to prove that each statement is true for every positive integer n. $\sum_{i = 1}^{n} 5 \cdot 6^{i} = 6 (6^{n} - 1)$