Use mathematical induction to prove that the statement is true...

Question

Use mathematical induction to prove that the statement is true for every positive integer n. 1 + 4 + 4^2 + ... + 4^(n-1) = ((4^n)-1)/3

Accepted Answer

Welcome back. I'm so glad you're here. We've got mathematical induction here. We want to go through the process of mathematical induction to prove that the following statement is true for each positive integer. And now this is quite a process, but we've got this. So the first step as we recall from previous lessons in mathematical induction is we need to show That as sub one is true. So you've got to prove that first. Okay, so we'll say that S sub one means we're going to plug a one in where there's an end. Okay, So we go here to the left hand side. So instead of three to the two end, it's going to be three to the two times one minus two equals. And then on the other side now, instead of an end, it's going to be nine to the one minus one, All divided by eight. Now let's do the work on the left hand side. So two times one is two minus two, so that's going to be zero. And we know by definition that anything raised to the zero front exponents. Well that's going to be one. So the left hand side is one now on the right hand side, we've got nine to the first, so that's nine minus one or eight On the numerator. So we've got 8/8. Well that's also one. So both sides are one, we did show That as someone is true for this statement. So step two for mathematical induction is we're going to assume that S sub K is true. And then I'll get to the then in a moment. So let's assume s sub K is true. So we're going to plug in a K. For every and now. So S sub K. And we're going to copy the whole thing this time we've got one plus nine plus nine squared plus nine cubed plus everything in between plus three raised to the two K minus two equals nine. Okay -1/8. And I want to highlight real quick that we can do a little bit of work up here with this. We've got two K minus two. Well, we could factor two out in front there, so the k minus one. So three raised to the two times K minus one. Well three square that's nine. So we could say nine raised to the k minus one. That's equivalent for that part. That will help us in a few minutes. Okay, So we're assuming that this is true. That's just what you have to do with mathematical induction. Excuse me. So we've assumed it's true. And then what we have to do is prove that S sub K plus one is true. We have to prove that. And that's where this one gets to be a little bit fun. So we're going to do s sub K plus one and we'll copy down that first part one plus nine plus nine squared plus nine cubed plus everything in between plus. Now, I'm going to take that top part that we rewrote and nine raised to the k minus one. That's that one plus. And now, instead of a K we're putting in a K plus one because this is the number that comes directly after K. Right? So nine to the K plus one minus one equals instead of nine to the cake. Now it's nine to the K plus one And -1. All over eight. Alright, here is the fun part. We know that this whole first chunk is the same as this whole first chunk up here, which is therefore the same as the right hand side of the equation here. And we can replace the whole chunk down here with the right hand side of the equation from up there, that makes this much prettier. So we'll say we have nine to the K. Not cave minus one, but it's nine to the k -1/8. And now bring down this part Plus 9 to the K-plus 1 - equals and then bring down the right hand side equals nine to the K plus one minus 1/8. Now it's really just algebra. So we're going to get rid of the eights, let's multiply everything by eight. So the eights will cancel out here And that's just going to be 9 to the K -1. And now that nine is going to be multiplied by an eight so we'll just keep it in parentheses for now. So eight times nine to the K. Okay Plus one minus one. That's just nothing. So now it's nine to the k equals nine to the k plus one minus one. And that eight also Gets canceled out by the eight were distributing. So now we get to rearrange this a little bit, let's take this minus one and we'll just move it to the end. We'll say nine K Plus eight times 9 to the K minus one equals nine to the K plus one minus one. And now those first two terms, they've got something in common, they've got a nine K. Nine to the case. So we'll factor that out. We've got nine to the K. Times while nine to the K divided by nine to the case one plus. And this other term is eight minus one equals nine to the k plus one minus one. And inside those parentheses one plus eight that equals nine. So we've got nine to the K. Times nine minus one equals nine to the K plus one minus one. And we can rethink this nine to the cape times nine. Will they have a common base? And this nine it's like nine to the first. So now we can combine them and say nine to the K plus one minus one equals nine to the K plus one minus one. They equal they match we have proved as to the K plus one is true. Which means that this statement is true for the values of N And we've done it well done, we'll catch you on the next one.

Use mathematical induction to prove that the statement is true for every positive integer n. 1 + 4 + 4^2 + ... + 4^(n-1) = ((4^n)-1)/3

Key Concepts

Mathematical Induction

Geometric Series

Formula for the Sum of a Geometric Series

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