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.

2 + 4 + 8 + ... + 2^n = 2^(n+1) - 2

Accepted Answer

Hello. Today we're going to be proving that the given statement is true for every positive integer. Using mathematical induction. So what we are given is five plus 25 plus 1, 25 plus all the terms to the end term five to the power of N. And this summation is represented by the statement five to the power of N plus one minus 5/4. Now, in order to prove that this is equal to the summation. The first step in mathematical induction is to show that this statement is at least equal to the first term and we can do that by allowing end to equal to one. So the first step in mathematical induction is to allow end to equal to one and set our statement equal to the first term of the summation. And doing this is going to give us five is equal to five to the power of n plus one, which is going to be one plus one because N is equal to one minus five. All of that over four. Now, five to the power of one plus one is going to give us five squared and five squared is going to give us 25. So we have five is equal to 25 minus five. Over 4. 25 minus five is going to give us 20. So five is equal to 20/4, which itself simplifies to five. So this at least shows that the given statement is equal to the first term in the some. Now we want to go ahead and assume that this statement here is equal to all of the terms in the summation. And so the second step in mathematical induction is to show that n is equal to K. When n is equal to K is going to give us the statement five plus 25 plus 1, 25 plus all the terms to the K term. Now five to the power of K is equal to five to the power of K plus one minus 5/4. And the purpose of this is to show that the statement is equal to all the terms in the summation. So for now we're going to assume that this is true. Now the next step is to show that this statement is equal to the next term in the summation. And the next term is when n is equal two K Plus one. So we're going to add the next term into our summation. This is going to give us the statement five plus 25 plus 1, 25 plus all the terms plus five to the power of K plus the next term now, which is going to be five to the power of K plus one. And now we need to just go ahead and make a slight adjustment to our original statement which is now going to be five to the power of K plus one plus one, which is going to be five to the power of K plus two minus 5/4. And now we want to prove that the left hand side is equal to the right hand side. Now this is the bit of a tricky part because how can we show that an unlimited amount of terms on the left hand side is equal to the statement on the right hand side? Well, notice one thing. This portion of the left hand side is the same as the statement in step two. So we can actually substitute this statement with five to the power of K plus one minus 5/4. And doing this is going to give us five to the power of K plus one minus 5/4 plus five to the power of K plus one. And that should equal to the right hand side statement, which is five to the power of K plus two minus 5/4. Now, at this point it's just going to be algebra. So for now let's go ahead and put five to the power of K plus 1/1. And we're going to want to add both of the fractions on the left hand side together in order to do this, we're going to first multiply this fraction by 4/4. That way we get another fraction with the same denominator of four and multiplying five to the power of K plus 1/1 by four or four is going to give us on the left hand side, Five to the power of K plus one minus 5/4 plus four times five to the power of K plus 1/4. And for now I'm going to label the right hand side as our H. S. On the left hand side, we can go ahead and combine both of the fractions together since we have the same denominator and I'm gonna go ahead and put the terms that have five to the power of K plus one together. So this is going to give us five to the power of K plus one plus four times five to the power of K plus one minus 5/4. And that should equal to our right hand side. Now on the left hand side, notice that the first two terms in the numerator have the same factor of five to the power of K plus one. So we can go ahead and factor out five to the power of K plus one from those first two terms. And that's going to give us five to the power of K plus one times one plus four minus five. All of that over four. Now you may be thinking to yourself, well wait a minute, where did this one come from? While in the previous step five to the power of K plus one has a coefficient of one. So when you factor out five to the power of K plus one from the first two terms, that's going to give you the coefficients one plus four and one plus four is just going to give us five. So what we have now is five times five to the power of K plus one minus 5/4. Now here five has the exponents of one. So we can go ahead and use exponential multiplication to simplify five times five to the power of K plus one. And doing this, all we need to do is just go ahead and add the exponents together. And once we do that, that's going to give us five to the power of K plus one plus one minus 5/4. And that should equal to the other statement which was five to the power of K plus two minus 5/4. Now on the left hand side, all we have to do is just simplify the exponents and K plus one plus one is going to give us k plus two. And that means that the left hand side of the statement is going to equal to the right hand side of the statement, which shows that the original statement given to us is equal to all the terms in the original statement. And with that being said, the answer to this problem is going to be d. So I hope this video helps 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. 2 + 4 + 8 + ... + 2^n = 2^(n+1) - 2

Watch next