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JH
Numerade Educator

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Problem 88 Hard Difficulty

The Fibonacci sequence was defined in Section 11.1 by the equations
$ f_1 = 1, f_2 = 1, f_n = f_{n -1} + f_{n - 2} n \ge 3 $
Show that each of the following statements is true.
(a) $ \frac {1}{f_{n - 1} f_{n + 1}} = \frac {1}{f_{n - 1} f_n} - \frac {1}{f_n f_{n + 1}} $
(b) $ \displaystyle \sum_{n = 2}^{\infty} \frac {1}{f_{n - 1} f_{n + 1}} = 1 $
(c) $ \displaystyle \sum_{n = 2}^{\infty} \frac {f_n}{f_{n -1} f_{n + 1}} = 2 $

Answer

(A). $=\frac{1}{f_{n-1} f_{n+1}}$
(B). $=\frac{1}{1 \cdot 1}-\frac{1}{\infty}=1-0=1$
(C). $=1+1-0-0=2$

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Video Transcript

the Fibonacci sequence is given below. So recall it's defined by starting out with one. Then you have one after that. And then the next term you obtained by adding the previous two terms. So the third term would be the sum of the previous two terms, which is too. Then you add the previous two previous to and so on. Okay. And so that's that's all right. There, That's it. Some of the sequence. So for part, a wealth of here We have three identities and hopefully we could use part eh? To do B and C. So first, let's start off with party that unless you show that this is true. So personally, I don't like those denominators. So I'll go ahead and multiply both sides by the denominator on the left just to get that out of there. Then on the left side, we just have one. And on the right hand side when we multiply and then simplify a little bit here. So let me show my work. So let me go a few steps back. Here we have. This is after multiplying both sides by I the denominator on the left. So then we have this. Cancel a CZ, much as we can. Here, cancel those F and minus ones F n plus one over here. And then we're left with F N plus one over FN minus F and minus one over FM. And so we could put these together here. So in this case, actually, let's just multiply both sides by FN first I'Ll just get that common denominator. So we have one over here this on the far right, multiply the FM to the other side and then finally add this blue term to the other side and then we have f and minus one plus f end equals f n plus one. And this is true based on the definition over here. So this is true True statement. These are both true in this verifies party, eh? So we can rewrite this fraction and party on the left using the decomposition on the right and then we use that in parts of being C. So for part B, let's go ahead and start on the next page. So Ron part B here we're something from two to infinity, one over F and minus one and plus one, and we won't would like to show that this equals one. So first, let's recall the formulas of this earned less issues party right now. So this is using partner right here. I'LL go ahead and replace the fraction with the become decomposition. So we have won over f n minus one f n minus one over f n f n plus one. This is just from party, eh? We just showed us a few moments ago. Now let's do the telescoping. So first I'll rewrite. This is a limit as que goes to infinity And then I'll look at the partial, some from some number from two up to some number. Okay, so what will evaluate the Somme and then at the very end will come in here and take the limit? So let's go ahead. And now this is where we're setting up the telescoping. So plug into first. So that's the first term when you plug in and equals two, then plug in and equals three and so on, and I will keep going. And then I get to the second to last term when I plug in K minus one for end, and then I would add the final term in When I plug in and equals K, This is one over F k minus one F K minus one over F K, then F K plus one. Now we could telescope in, cancel as much as we can, so we see that the first term will never cancel because none of the other denominators have f one. However, after you have three, we have cancellation and will continue to have cancelation all the way up until we get to this K minus one f k. But then we're left with one more term out here that will not cancel. So the next thing to do here is just to evaluate this. So let's just go ahead and write this on the next page and then take that women. So I'm out a room here, so I have one over F one F two minus one over f k f K plus one, and we're taking the limit as cables to infinity. Now this cake goes to infinity. This is just the cost in here, and we know that F one equals one and F two equals one. So we just have one over one there. But since these terms are both going to infinity and there in the denominator. So we're using the fact here that the feminine cheese do numbers do go to infinity. The denominator goes to infinity so the fraction as a whole is closer to zero. Therefore, the entire sum is just one. And that is exactly what we wanted to show for part B. We wanted to show that the sun was equal to one now finally was also look at party. So let me go to the next patient for part. See, we are looking at the Somme. Very similar looking some. But this time we have FN in the numerator. And we'd like to simplify this and eventually show that this should equal to and we're done. So here we cannot try to use the formula that we had from part A. So from party we had. So then Now. But we also have this FN in the numerator now. Oh, so luscious introduced that fn over here by multiplying it up, Doc. So this is what we have on the left is the same as the fraction over here on the left. And then after we could simplify over here, cancel those F ends we have won over F and minus one minus one over and plus one. And so this is what we can do when we replace the sum here. So now we have the sun going from n equals to to infinity. And now we have won over F and minus one minus one over F and plus one. So that's just using part A Once again, As I mentioned, we were going to use party for part B and C So now we'LL set up the telescope ing again. So the telescope So first I'll rewrite that infinite sum is a limit. And then instead of doing the whole sum, I just look at the partial some and then I just keep this the same. Here's my a n Now let's expand this some and then we'll do our telescope So start that, adding by plugging in and equals two. So one over F one minus one over at three and then one over F two minus one over a four plus one over a three minus one over a five and so on. And then we wouldn't have to also add the last few terms so I had three more terms and over here on the on the right end. So the very last term is when you plug in K. Before that, you plug in came on this one and before that you plug in K minus two. And so here was our formula, so we would have minus one there. So then let's try to see if we could cancel out as much as possible. So f one enough to these won't cancel because they're the negative star F three. So we'LL still have won over F one and we'LL still have won over F too. However, once we get a half three, we see that we'LL start having some cancellation here that'LL cancel with that and this negative one for will also cancel if I act. If I keep adding terms on the other hand, let's look on the other side that far right This term will not be canceled out because if you look at the positive fractions on the left of all the front disease, the farthest they go up to its K minus one, so we'LL be able to cancel the these out up to K minus one, but unfortunately, the minus one over F K and won over F K plus one. Those will still appear. So let's go to the next page to write this out. And we're still taking the limit as Kay goes to infinity. So now let's take that limit. Since, as we noted in part B, f K goes to infinity, the NRC numbers do go to continued to grow without bound. That means that these denominators both go to infinity, so the fractions is a home to go to zero in the limit. Burgess left over with one over F one plus one over half, too, but recalled by definition, F one F two are both equal someone, So we just have one plus one equals two. And that's exactly the sum that we wanted for parts he. So there's our final answer.