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 + 2 + 2^2 + ... + 2^(n-1) = 2^n - 1

Accepted Answer

Hello. Today we're going to be proving that the given statement is true for every integer using mathematical induction. So what we are given is the summation one plus three plus three squared plus all of the terms to the term three to the power of n minus one. And the statement that represents the summation is three to the power of n minus one, divided by two. So how do we prove that? This statement is equal to this summation. While through mathematical induction we can show on the first step that this statement is at least equal to the first number in our summation. So in order to show that the statement is equal to the first term, we want to go ahead and let N equal to one And letting n equal to one is going to give us one is equal to three to the power of N which is three to the power of 1 -1 divided by two. Now three to the power of one is just going to give us three and three minus one is going to give us two. So what we have is one is equal to 2/2, which simplifies to one. And that shows that the original statement is at least equal to the first term in the summation. Now we want to make a statement that shows that the original statement is equal to every term in the in the original statement. And the second step of mathematical induction states to allow N two equal to K. That way we can come up with a statement that shows that the original statement on the right hand side is equal to every term in the summation. So letting an equal to K is going to be one plus three plus three squared plus all of the terms to the case term now which is three to the power of k minus one. And that is going to equal to 3 to the power of K -1/2. Now, for now we're going to assume that this statement is true and we want to come up with another statement that shows that the original statement is going to be equal to the next term in the summation. In order to show that mathematical induction states that we need to add the term when n is equal to K plus one. So adding the next term is going to give us one plus three plus three squared plus all of the terms to three to the power of k minus one. And now we're going to add the next term, which is going to be three to the power of K plus one minus one. Because all we're doing is adding adding +12 K. And this is going to equal to Now the adjustment with K +13 to the power of K plus one minus 1/2. And from this point it's just going to be algebra because now we want to let the left hand side equal to the right hand side. But how do we do that when we don't know the amount of terms given to us on the left hand side. Well, here's the tricky part notice that this portion of the statement is exactly equal to the previous statement we came up with with that being said, we can allow this portion of the, of the summation to equal to three to the power of cain minus 1/2. So once we make that substitution, we're going to get three to the power of K minus 1/2 plus three to the power of K plus one minus one, which is just going to be three to the power of K. And that is going to equal to three to the power of K plus one minus 1/2. Now we want to show that the left hand side is going to equal to the right hand side. So let's go ahead and start that process. For now I'm gonna go ahead and put three to the power of K over one and I want to go ahead and combine both of these fractions together in order to do that, I need to go ahead and multiply three to the power of K over one by 2/2. That way I get a fraction with the same denominator of two and doing this is going to give me three to the power of cain minus 1/ Plus two times three to the power of cane over two and that's supposed to equal to our right hand side statement. Now I can go ahead and combine these fractions together and I'm gonna go ahead and put the terms involving three to the power of K in front of the numerator. So doing this is going to give me three to the power of K plus two times three to the power of k minus 1/2. And that should equal to our right hand side statement. Now notice that the first two terms in the numerator have a common factor of three to the power of K. So we can go ahead and factor out three to the power of K. From the first two terms in the numerator and that's going to give us three to the power of K times the quantity one plus two minus one, all of that over two. And that should equal to our right hand side statement. Now you may be thinking to yourself where did the one come from? While in the previous step three to the power of K has a coefficient of one. So when you factor out three to the power of K from the first two terms you're going to add the remaining coefficient which is going to be one plus two. So adding the remaining coefficients is going to give us three times three to the power of k minus 1/2 and that should equal to our right hand side and for this product right here we have three times three to the power of K. Now three is going to have an exponent of one. So we can combine these terms together using our exponential multiplication and that states that all we have to do is add the exponents together for any values that have the same base. So we can simplify this quantity to be three to the power of K plus one minus 1/2. And that is equal to our original statement, which is three to the power of K plus one minus 1/2. And since this statement is true, their mathematical induction, this shows that the original statement is equal to all the positive integers for our original statement. 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. 1 + 2 + 2^2 + ... + 2^(n-1) = 2^n - 1

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