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13_ The infinite series 2 ax has nth partial sum S = (-1)" for n 2 1 _ What is the sum of the K]series 2 ak ? k-L...

Question

13_ The infinite series 2 ax has nth partial sum S = (-1)" for n 2 1 _ What is the sum of the K]series 2 ak ? k-L

13_ The infinite series 2 ax has nth partial sum S = (-1)" for n 2 1 _ What is the sum of the K] series 2 ak ? k-L



Answers

Find the sum of the finite geometric series. $\sum_{k=0}^{13}\left(\frac{1}{2}\right)^{k}$

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So in this problem we're being asked to find the some of this geometric series. Well in order to do this, we're going to use our some formula which says that as a Ben. There we go. I got to have the market going. So S. F. N. Is equal to a someone times one minus are to the end, all over one minus R. So the first thing we need to find is a someone to do this. We substituted zero for K. So we have one half raised to the zero power. Well anything to the zero power is one next. We have to find our will to find our, we need the second term in this series in the series. So to do this will substitute and one for K. So one half to the first power is just one. So now we can find our we have to figure out what and I apologize to shit on one half. So the fines are we figure out, well one times what is one half, Well one times one half is one half, so R. Is equal to one half. Lastly we need to find end. Well typically are lower limit would start with one. If it did we would have 13 terms. But because we subtract that by one, that means there is an additional term meaning that N is equal to 14. So now we'll substitute these values into our formula. So have s of 14 equal to one times 1 -1 half, raised to the 14th power, All getting divided by 1 -1 house. You might want to put the one happening premises. Now the nice thing is we can do this right on our calculator so you can type the whole numerator into your calculator once, press center And then our denominator 1 -1 half is equal to 1/2. So you can just divide it by .5. And so then when you turn it into a fraction, you will get 16,383 divided by 8,192. So that would be the some of the series.

Okay, so let's start by shifting or index. Since we want this to start at one on this to be a value minus one. So it's a end. People to K, um, plus One and Alvin K is equal two and minus one. Okay, so we have some from and it's equal to one. And then we have two K's now and minus one, and then our last time. 13. Well, that's equal to and is equal to r K value. That's 13 plus one. That's not fortune that So now let's find Rs of fortune that's equal to our first term or notice. That's our first term is one. So this is our A one value. And this, too, is our our value. So we have one times one minus to the power of 14/1 minus two. So it's put back in two more cooperative, and we get this is equal to 16383

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