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(A) $2^{n}-n-frac{1}{2}$(B) $1-2^{-n}$(C) $n+2^{-n}-1$(D) $frac{1}{2}left(2^{n}-1ight)$...

Question

(A) $2^{n}-n-frac{1}{2}$(B) $1-2^{-n}$(C) $n+2^{-n}-1$(D) $frac{1}{2}left(2^{n}-1ight)$

(A) $2^{n}-n-frac{1}{2}$ (B) $1-2^{-n}$ (C) $n+2^{-n}-1$ (D) $frac{1}{2}left(2^{n}-1 ight)$



Answers

$\frac{e^{-s}\left(3 s^{2}-s+2\right)}{(s-1)\left(s^{2}+1\right)}$

This question says to use sigma notation to write for some. So you have your sigma and you really only have a beginning term and you're ending term to go on. They didn't give you anything in between. So if you kind of look at what's changing as you go from the first term to the last term, you've got to over an which is two men over end. So you can kind of see that this part up here is changing. And then you've got this minus two over an and it goes to minus to an over in so you can see that that numerator is changing. And so what you notice is not changing. Is this too over in and two of her end So we can kind of put that out here and not included in our summation. It wouldn't really matter if you included it or not. We need to pick an index estimation. I'll just pick a for this one. So where does case start and where does he end? So he's would be starting at a value of one and ending at a value of and because if you use your last term here, which, when there's variables involved, it's common to use the last time you can kind of see that you're plugging in something in that numerator. So if you think about this, you would have to win over an. So that's your two K over an and all that being cubed minus and then you could see that you had to end over end. So your index is changing, and that would be two K over end, and that would be your expression to write your son.

So they have us give us some hooch. We only have two terms. However, the terms of the beginning of the end. So hopefully that doesn't know how do thems one plus three over in on, then squared out. And then these multiply that by three over in. Then you had a somewhat of a bunch of terms, and then you end up with the term two domes, one plus three or and square about a squirt on the bottoms three or in. Okay, So our first thing here it seems that from off the song, well, we better begin factories there while the two is always there. Two barley weekend also factored in. So it should be selfie over the farm. Two times, three over. And I did Thumbs are so from Michael's somewhere up to somewhere something. So, uh, we're going to identify what is in there. So there is a one in there, so it seems to be common one within plus something over, right. Is you conceding those two terms So should be one plus something or end and then squaring that But also well, you see here that there is also have three things are going to be feeder then. Or these three can be seen us three times one, presumably in these next place you how you're gonna have. They're among the form one plus the them's too over that squared. All these lee multiplied by the so on the three or end. Uh, so personally, you'd have the temps. I weren't. So what do you in Dublin until? Well, old that he said, Well, you still I goes one me and double I goes in. Yeah. So, um oh, that taken into account should be reading us. Do three n with three or four or six, if you will. I'm being these terms names one plus the time site that we're in of these hoping squared the wall, He seems Have you started Michael's one? And you end up that I close in. So it seems to be that

Okay, We're gonna use the racial, their route. This to check if the Siri's converges off diverges. So we have to calculate the limit when ghost infinity off the en route. Oh, and the one of grand minus one to the end. Now it's easy to see the the improved will cancel with in power. And we have left and over one over, then my nose once. Now, when it goes to affinity, this was to infinity, and this goes to see a row. So we will like to apply something like no Peter to be able to solve that. But before that, we're gonna use this common technique, the consistent applying e and Ellen. We're taking away the one from the limit because we know he's just once. So let's complete the limit off Ellen off into the one of the end within this because, uh, one over then in explosion will fall in front off the island. So now we have one over in a line open. Now, when it goes to infinity, that goes to infinity and and goes to infinity too. So we're gonna ply Lopata for that. We need a continuous variable. So we changed the end by X, and we shot the derivative off. Ellen, that is one Rex and then everybody fixed. That is one. So when exposed to infinity, this goes to syrup. Now we go back to the original limit and we have this expression so we can replace the limit by what we just fine. Now that is, eat to the syrup minus one. Now you 20 It's 11 minus one is zero. This is the smaller than one. And so the route does tell us that the serious convergence

This question. We're going to multiply the monem. You look to end to this trying mobile, foreign squared, minus foreign plus one. To do that, we're gonna use the distributive property. I'm gonna first multiply the to end to the foreign squared. That gives us eight n to the third power. Using the product rule, we can add the exponents on the ends. Now I'm gonna multiply to the next term to end times negative foreign. That gives me negative eight end squared. All that these ends have a power of one and one plus one is two. Finally, in two end times, our last one of one just gives us to end. And that's our final answer. There's no like terms in here, and this cannot be simplified further.


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