Is gonna be that times, is gonna be the second term, the third term is the The previous term times 9/10, so it's gonna be 10 times nine over 10, and then the next term So the first term is 10,Īnd then the next term, so the second term A sub two is equal to A sub one times 9/10, alright. Series defined recursively and so it's useful to just think about what it would actually look like. Give you the exact value when you take something to the 80th power, but this is what that sum is going to be. Value or close value actually, most calculators don't Simplify it more but, if we had a calculator we could actually find this exact Subtracting a negative that's just gonna be adding the positive, so all of that over 1.99, and we could attempt to So that's going to be the same thing as 0.99 to the 80th power, and all of that over, well Well, we're taking it to an even power, so it's going to be positive, There to make sure we are taking the negativeĠ.99 to the 80th power. Worry too much about that, and so this is going to be one minus, so negative 0.99 to the 80th power, I should put parenthesis This a little bit, this is all going to be equal to, oh that one we don't have to To be to the 80th power, all over one minus negative 0.99. Now to go to the next term, we're again multiplying by negative 0.99, so the common ratio is not positive 0.99, but negative zero, negative 0.99, so let me write that, negative 0.99, and of course that is going Term to the second term, we multiply by negative 0.99. Well to go from term to the next what are we multiplying by? Well to go from the first ![]() So at first you might say well maybe the common ratio here is 0.99, but notice we have a change in sign here, and the key thing is to say Ratio to the 80th power, to the 80th power, all over,Īnd I'm leaving a blank because we still need toįigure out our common ratio, all over one minus our common ratio. So this is gonna be equal to our first term is gonna be one times one minus our common Right over here is the 80th, the 80th term, 80th term. Taking it to the third power so on and so forth, so this Taking it to the second power, the fourth term is where we're Taking it to the first power, the third term is where we're When we're taking things to the zeroth power, we're Here", but be very careful, because the first term is You might be tempted to say "Okay I'm gonna take it to the 79th power, "there must be 79 terms Think about how many terms we're gonna take the sum of. ![]() Simplify this even more, but this gets up pretty far, at this point it is just arithmetic. ![]() ![]() So one minus 10/11 to the 50th power over, this is 11/11 minus 10/11 is one over 11, and so this is the same thing as multiplying the numerator by 11, and so this is gonna be equal to 11 times one minus 10/11 to the 50thpower, and you can try to Here just to make sure we're not just takingġ0 to the 50th power. And so I'm not gonna solve it completely, but we can simplify this a little bit, this is gonna be one minus, let me put parenthesis So we can apply the formula we derived for the sum of a finite geometric series and that tells us that the sum of, let's say in this case the first 50 terms, actually let me do it down here, so the sum of the first 50 terms is going to be equal to theįirst term, which is one, so it's gonna be one times one minus, let me do that in a different color, one times one minus the common ratio, so the common ratio here isġ0/11, 10/11 to the 50th power, to the power of how many terms we have, all of that over one The next, what are we doing? Well, we're multiplying byġ0/11, to go from one to 10/11, you multiply by 10 over 11, then you multiply by 10 over 11 again, and we keep doing this, and we wanna find theįirst 50 terms of it. We're asked to find the sum of the first 50 terms of this series, and you might immediately recognize it is a geometric series. In summary: finite/infinite refers to the number of terms of the series, but convergent/divergent refers to the type of limiting behavior of the sequence of partial (finite) sums for an infinite series. If this limit of partial sums exists (which occurs when either a = 0 or |r| = 1), then the infinite geometric series is divergent and the sum does not exist. Note that the sum of this type of series is defined as the limit of the sequence of partial sums (that is, sums of finite series of the form sum n=0 to k of ar^n) as the number of terms approaches infinity. For this type of series, the question of convergence vs. Note that this type of series has infinitely many terms. Note that this type of series has finitely many terms, and so the sum always exists.Īn infinite geometric series is a series of the form sum n=0 to infinity of ar^n. A finite geometric series is a series of the form sum n=0 to k of ar^n.
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