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JH
Numerade Educator

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Problem 90 Hard Difficulty

(a) A sequence $ \left\{ a_n \right\} $ is defined recursively by the equation $ a_n = \frac {1}{2} \left(a_{n - 1} + a_{n - 2} \right) $ for $ n \ge 3, $ where $ a_1 $ and $ a_2 $ can be any real numbers. Experiment with various values of $ a_1 $ and $ a_2 $ and use your calculator to guess the limit of the sequence.
(b) Find $ \lim_{n \to \infty} a_n $ in terms of $ a_1 $ and $ a_2 $ by expressing $ a_{n + 1} - a_n $ in terms of $ a_2 - a_1 $ and summing a series.

Answer

(A). Seems to approach $\frac{2}{3}$ of the way from al to a2
(B). $\frac{a_{1}}{3}+\frac{2 a_{2}}{3}$

Discussion

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Video Transcript

so a sequence a n is the fine recursive Li Bai this formula up here, assuming that is larger than three, three or more, a one and a two, or any numbers that we choose here. Now let's experiment with some values of a one and a through that we get to choose and then, well, guess the limit. So for part A, let's come over here and choose some values for zero on a one. This is our A zero appear F zero, and this is our A one half of one. So let's do a zero equals zero. A one equals ten and then our guests for the limit. Here we can draw, like, say, if we look at the tenth term in the sequence, it's given down here by about point six six, and you could kind of see that the numbers were getting. It looks like they're hanging up at around point six six, so that would be a guest here. Six and two thirds. Now let's maybe try different values of of a one and a two. So here, let's try, for example, negative ten and then positive ten. And then here maybe we'LL need more terms to see what's going on here. Let's try maybe thirty terms and it looks like they're going to about three and a third. So here I should have done a one and a two originally. Sorry about that. Then, down here, a one is negative. Ten. A two equals ten. Then I could see that the limit is about three and a third action. So now let's go ahead apart. Be onto the next page. So for part B, let's record some values of end. Hey En. And then also the A N plus one minus an eye or excuse me, I should do a M minus and minus one. So, for example, if N is one a one, nothing yet to subtract. Two, then a two you can't use the Rikers in yet. Remember, the formula starts for and bigger than other people's two three. That's why I hear I'm not using the formula. And then over here I have a two minus a one. Now I started using the formula, So instead of a three, I write anyone plus a two. These are both over, too, and now I have a three minus eighty two. So in this case, I have a one over two minus a to over to after I replace the three with this and then subtract up too, to do one more value here. Now have a four a four. I could go ahead and use the formula. So a four is a three plus a two over, too. And that I know a three from over here. So after simplifying all of that and writing it in terms of a one and a two and then similarly so this Let's see here. Gotta be careful with a one minus a one over four plus a two over four. Sorry about that. And we could keep going, and that's better. But from this all right going to conclude is that we can rewrite the limit of a M as n goes to infinity. Well, now we just look at the right most calm and you could see that if we just add up these terms, we'Ll end up with a N because here will have the cancellation basically a telescoping some So one will die if we take the limit of this. And after we do the telescoping we'LL end up canceling the A one. Well, they won. Will that part? We actually have to make up for that by just writing a one here because we see that doesn't cancel. And then we'LL have negative a one plus a, too, so that a ones will cancel. And then we would do negative a two plus a three day two's would cancel, and you could see if we keep going in this pattern, everything cancels out except the end. So here we're cancel that a one's a two's, a threes and so on and your leftover with a Now we'LL use the fact that we've in the apprentices. We've simplified all of these expressions in terms of a one and a two on Lee, so going on to the next page, we have a one. Plus, these first two terms will remain because they're already in terms of a one and a two, and then the next one. We already worked this out on the table on the previous page, and I could keep going in this pattern. You could see that it's a geometric series, so the first term will leave that alone. Then, after that, we can rewrite out this sum and two minus a one. What? We'LL have to be careful here because this looks we have this geometric But we also have were multiplying by negative because you could see the role of the negative. Here, a one is negative. So it starts off positive than negative, positive, negative and so on. So that's our theories. This is a geometric series. Our equals negative one half. Luke's gonna sloppy there. So we know the formula for geometric series. So this term just comes down a one. Then you would do the first term of the series by plugging in I equals zero. That's just a two minus a one. And then you divide by one minus R And then let's just go ahead and clean up this fraction and we'LL end up with a one over three plus two es tu all divided by three. And that's our final answer for Barbie