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Determine which of the geometric series is convergent and find the sum_(-1)"-13n Er=1 4n-L([n2)}(c) Enzo3.2-n...

Question

Determine which of the geometric series is convergent and find the sum_(-1)"-13n Er=1 4n-L([n2)}(c) Enzo3.2-n

Determine which of the geometric series is convergent and find the sum_ (-1)"-13n Er=1 4n-L ([n2)} (c) Enzo3.2-n



Answers

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
$ \displaystyle \sum_{n = 0}^{\infty} \frac {3^{n + 1}}{(-2)^n} $

Let's determine whether this geometric Siri's converges or diverges and if it converges, will go ahead and find the sun. So first, let's recall Geometric series is typically written in the following form and we know that this will converge on ly when I flew. Value are is less than one, otherwise it averages so otherwise. But I mean by that is the absolute value bar is bigger than or equal to one and in our case, in diverges So this is for geometric So first, let's go ahead and rewrite are some so that it looks in this form so we could say what our is. So what I'll do there is I'm just going to play around with the numerator I like a three two that I prefer three did and power are negative three then power So what I'LL do here is I'Ll just multiply divined by negative three That gives me negative three the end over negative three So I have negative three to the end over negative three and the numerator and I'm still dividing by four to the end and then here because these air both to the end power let's go ahead and pull out the fraction. So that would be negative. Three over four to the end and then were multiplying this by negative one over three. So we can see that are are the thing that's being raised to the power equals negative three over four. It satisfies this which is less than one so and is conversion and purchase. Using this fact appear converges when the absolute value are is less than one three over for less than one so converges. And in this case, we also have a formula for the sun. So let's write that out and equals one to infinity. Negative three over four to the end. Negative one, sir. So the formula says you take the first term and then you divide by one minus R. So in our case, the first term is the one that you get by playing in the first value, and down here in our case happens to be one. So plug in one friend. So you get negative on the top. You get negative three to their one that'LL cancel with this minus three. So you just get one over four in the dim writer when you plug in a and equals one. And then you have one minus our value of our, which was negative, three over four. So this right here, we can go in and simplify that. Let's just go to the next page to write this. So we had one and they canceled that double minus. So that's one over four. And then we have seven over four. Selection's cancel most force to get our final answer of one over seven. So the Siri's conversions and won over seven is the value of the Siri's.

Around. My name is Kevin Chirac. Let's take a look at the infinite Siri's and equals one to infinity. But this time it's gonna be defined by its notation. So being negative three to the end, minus one native three is going to in parentheses, so it's gonna create an alternating Siri's and it's gonna be over four to the end. Now the first thing eyes to figure out whether or not this is a geometric series or this geometric serious. We do know that its geometric Siri's and we can actually see that by doing a little bit of simplification here, where we can essentially factor out this negative one to the end, minus one power. I don't want to say equals doesn't want to carry through that notation again, but we could essentially factor out that negative went to the an minus one power. We then have three to the end, minus one power. That is what it was being multiplied by. And then we have that is, over four to the four times four to the n minus. When I could write it this way, well, all of this now can get rewritten as that negative one which means it's an alternating geometric. Syria's times 3/4 to our end, minus one. And then all of this is going to be times 1/4. So all of this could be substituted for the inside of our summation notation here, or what's gonna correspond well with us. Calculating it is for us to squish it back together just a bit and to say negative 3/4 to the n minus one times 1/4. So as we calculator geometric Siri's, we know that the summation is going to be equal. Teoh a over one minus are we can see from the form that we wrote above that are is going to be negative 3/4 and I have one minus a negative 3/4. And what about that? A. Will be that first value? Well, that first value occurs with the first entry into our Siri's, which would be when N is equal to one. So if we put in and is equal to one into all of this, then we're going to have the entire first term is going to know out to be just one, and it's three times 1/4. We know the numerator here is going to be 1/4 so let's begin our simplification

And n equals two. And here it is infinity minus, went to the bar and upon hold the power to end As equals -1 to the power to born to the power for plus minus went to the power three, a pound to the power six plus one upon To power eight plus so on. Here Is one upon to the power four & R is equal to -1 upon ford, therefore is infinitely equals Airborne, 1- are therefore it is equal to one upon tool. The power four upon 1 -1. This equals one upon 16, divided by one plus one upon four. This equals one upon 16, divided by 5.4, which is equal to one upon 16 into 4.5. Therefore this is equal to one upon 20 and this is the answer.

Let's determine whether this geometric Siri's converges or diverges and if it's come urgent will go ahead and find the summer as well. So geometric series is usually ran in the form you have, like some number A here, and then you have some term R Let's constantly being well supplied. And so the Siri's could start at any number of many times. Like in our problem, it's one and it goes up to infinity. So let's just rewrite our problems so that it looks like this. So I have six here and let me rewrite this. We have two of the two groups two and minus one. So this is to the two and over to to the one which is to square to the end over, too. So let's write that here we have two square and over to, and then we're still dividing by three end. Let's simplify six over to that's history and then combining these two terms right here we have for over three to the end. So now this were in this form here, and we see that are equals four or three a equals three. However, we know that a geometric series so geometric. Siri's converges on ly if the absolute value of our is less than one. But in our case, since the absolute value are equals for over three is bigger than one, the Siri's will diverge, so the Siri's is diversion, and that's your final answer.


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