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For the following telescoping series, find the value of the series.Xi [2 - 01]...

Question

For the following telescoping series, find the value of the series.Xi [2 - 01]

For the following telescoping series, find the value of the series. Xi [2 - 01]



Answers

Find the sum for each series. $$\sum_{i=1}^{20} \frac{1}{2}$$

We want to find the values of X for which the given series converges. The series is the sum from n equals 02 infinity of x minus two to the power of end This series is actually the form of a geometric series for which we can use the three steps I listen to the bottom of this page to solve this first. Let's define what a geometric series is. To make sure we understand the question properly. Remember that the geometric series is always in the form some N equals 100 T A R D n minus one or some and equal 02 infinity er to the end, it evaluates to a over one minutes are if it converges for absolute value are less than one. So if we first realize our series in the proper form, we'll evaluate where absolute value bars less than one in terms of X. And then we'll evaluate the limit. So we see that our series is already at the correct form, some N equals zero and x minus to the end, so this is A equals one, R equals x minus two. Perhaps the value of our less than one. We have absolute value of X wants to less than one. This gives X between one and three. So we can evaluate the limit A over one minus R equals 1/3 minus x minus two, which gives 1/3 minus x.

For us to find these something from sorry J equals 1 to 11 of minus one of the J. If you look back in the last problem, we know from there that we ended up with minus one and then plus minus one square to that minutes. Positive one. And as we went through, that gave zero up to 10. So as I write these, you can sort of see that every time there's a set of two, the odd power is gonna be negative and the even power is going to be positive. So it's gonna end up making it so that we have one of a minus and one of a plus, and they're gonna cancel. So these all and again, you can look back to the previous problem to sort of see what I'm talking about. A little bit more. These all go away. They cancel every set of two because one's negative ones posits. So the answer is just minus one of the 11 which is minus one

To find out what this sums to. Of course, this is infinite. So it would keep going, Um, we could convert thes two decimals. And so basically, this is adding 1/10 as the next term. And since the first one just starts out at one, we know that the index would start at zero. And then this could be written as 1/10 to the power. And because if Ana zero would get this if n is one we get, that end is to we get that and so on, down the down the line. And so this is really going to some 21 divided by one minus what's on the inside here, which is 1/10. And then that would end up giving us one divided by 9/10 and then flipping this, rearranging it, then this is just the same thing as 10 nights. So therefore, this whole some or the one above is going to some to 10 nuts

We want to find the values of X for which the given series converges. The series in question is a sum from n equals one to infinity of two to the end times X minus one to the end. So this series is actually in the form of a geometric series. We can use the three steps I listen to the bottom this slide to solve. But first let's define what a geometric series is. So we know how to execute these steps properly. Remember that a geometric series is always in the form some N equals one to infinity of a. R to the n minus one or some and equal 02 infinity of er to the end. A geometric series will converge to value a over one minus are if and only if it has absolute value are less than one. So we want to rewrite this in the term of a geometric series or rather in the correct form listed above and then evaluate where it converges. So we can rewrite are some for two to the N x minus one. At the end, the sum from n equals one to infinity of two times x minus one times two times x minus one of the n minus first power. Notice how he pulled out a power of two and x minus one, so with equals two times x minus one, R equals two times x minus one. So that means that absolutely you are less than one, gives absolute value two times x minus one, less than one. Or excess between one half and three half of the series converge. It converges to limit a over one minus R equals two times x minus 1/3 minutes to x equals two x minus 2/3 minus two X.


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