This problem. We want to find the 1st 10 terms of this sequence. Um, the sequence that's given to us here. There's an expression at Ben. It was won over Route five and then got that complicated bracket, um, so were asked to find the 1st 10 terms and then compare them to the Fibonacci numbers. So let's find the first couple of terms and see if we can generate a pattern. First term would be f of one F one that would be one over Route five one plus Route five to the power of one minus one minus Route five to the power of 1/2 to the power of one. And ah, we do have some like terms on the top one. Minus one would cancel Route five. Plus Route five would give us to Route five over, too, and that works out to one F of two is done in a similar manner that would be 1/5 and then one plus roof. I've squared minus one minus five, squared over two squared equals one over Route five, and we would expand that That would be one plus to re five plus five, minus one minus two or five plus five all over for and on the top we could simplify. We would have one plus five minus one minus five. So the constants go away. Two or five plus to revive would be four Route five over four, and that also works upto one good f of three. That would be one over route five. And then we got ah one plus 35 to the power of three minus one minus 35 to the power of three. All over two to the three, which is eight. So, yes, we're gonna have to expand some cubic there. That's Ah, unfortunate. So let's expand one plus Route five over here on the right side, one plus route five. So the power of three is equal toe one plus for five times one plus 15 times one plus Route five. 1st 2 would combine toe one plus two, Route five plus five. And, uh, we can combine like terms or six plus 25 And now we can multiply it one more time. That makes six plus 65 plus two Route five plus two times five, which is 10. So 16 plus eight, Route five. That's the expansion of the cubic. And if we actually did one minus Route five Cube that actually works out to 16 minus five. That's by a similar process. So carrying those two answers forward here, we would have Ah, 16 plus eight five minus 16 minus negatives of plus eight, Route five over eight sixteen's cancel. We have Ah, 16 Route five on top and eight, Route five on the bottom, which were accepted, too. So we've got one one. And to, um, you can continue to do F four F five F six. But the question was asking us to compare this to the Fibonacci numbers. It turns out these three terms here are the Fibonacci numbers, so you can verify them yourselves. But if we had continued this process, they would be 112 three, five, eight, 13 21 34. 123456789 and one more. 55. Those are the 1st 10 terms, um, of, uh, and using that formula Part B. That's just a show that's ah three plus or minus. Route five is equal to one plus or minus five squared over two. So again, it's a show. That problem. We want to start with the left side and arrive at the answer on the right, so that's a plus minus. So let's do it in two cases. Eso Let's do case one here. That's just work with three plus Route five. It's a three plus route. Five. Uh, you want to make it look like a fraction. So we want that over to there that would be written as six plus two. Route five over, too. And, ah, recognizing that we can turn that into a perfect square, try no meal by saying one plus 25 plus five over to that turns it into one plus Route five over too. Case, too, would be starting with three minus 35 2nd case that would be similar. That would be the same as six minus two Route five over, too, which is the same as one plus one minus to five, plus 5/2. And that is a perfect square. Try no meal, so that's one minus five over, too. So seeing that both cases work here just corresponds to the plus minus side, we can conclude that three plus or minus Route five is equal to one plus or minus five over two. Squared all over, too. There's the answer to part beat part C, and this is the hard part. It says. Use the result in part B to verify that F n satisfies the recursive definition of the Fibonacci sequence. Um, so the recursive definition is f m equals the previous two terms of F N minus one plus F and minus two. Ah, and the problem. We're given that definition so we can replace ah, the ends accordingly. So this is the same as one over route five times one plus route five to the power of N minus one minus or five to the power of all over two to the end equals won over five and then one plus five to the N minus one minus one minus route five to the N minus one all over two to the n minus one plus one over route five times one plus 15 to the N minus two, minus one minus Route five to the N minus two all over to to the n minus two. So that's using the definition and, ah, from here, we can see that, uh, one over Route five has in common with all the terms, so we can actually multiply everything by Route five. And those would go away. That means we just have those big brackets left. One plus 35 to the end, minus one minus 15 So the end all over to to the end equals one plus five to the N minus one minus one minus 35 to the n minus one. All over to to the n minus one. Less one plus five to the N minus two, minus one minus 15 to the N minus two over two to the n minus two. Uh, the questions asking us to verify this. So we just need to find a result that makes the left side equal to the right side. Obviously, we can't say that the two sides are equal right now because that they look different. How can we just say that it's equal to each other? Um, well, you would need to find a way to simplify this, and we need to simplify and using part B. So, um, one of the things that's bothering me right now is the fraction. So I'm actually gonna multiply all these terms by two. To the that'll get rid of all the fractions. So I'm going to say multiply by two to the So when we do that, we have one plus five to the power event, minus one minus 35 to the power of an equals. Ah, to be an extra tube. So one plus revive to the power of then minus one minus one minus revived to the n minus one. And we got two extra twos here. So four and then one plus route five to the head minus two minus one minus with five to the n minus two. Okay, um, next I'm gonna do Ah, let statement to simplify this because there's a lot of buying all meals here, and it's getting kind of messy, so we'll let a B one plus Route five and let B B one minus roof. I've just so things look a little bit more manageable so that a brook above equation would be written as a to the end minus B to the end equals to a then minus one minus two B and minus one plus or a and minus two minus for B to the N minus two. So I'm gonna group the A's and B's together. Ah, we got a to the N minus two age of the N minus one minus four A and minus two equals B to the end, minus to be in minus one minus four B and minus two. Uh, the goal here is the group the like terms so that I don't have to Ah, look at the one plus 35 and one minus revival on the same side of the equation. Now, um, I'm going to take out a common factor on each site. So on the left side, I could take out a common factor of A to the N minus two. So any to the n minus two that would give me a squared minus to a minus four equals same thing on the right side. B to the n minus two. Um, B squared minus two B minus four. So we know what a CZ and bees are. We couldn't fill that in. If we want. This would be the same as one. Plus Route five to the N minus two and then one plus roof. I've squared minus two times one plus. Route five minus four equals one minus through it. Five to the N, minus two and one minus. Roof five square minus two times one minus five minus four. Our goal was to use the result from part B. And I'm gonna write that here on the side. Part B said three plus or minus. Route five is equal to one plus or minus. Route five squared over too. Um So our goal is to use this result. I have the one plus five squared and the one minus five squared in my equation already. But I don't have that over to. So to create and over to I would divide this entire expression by two. I can't just change this entire thing. I would put a put a two in front as well. Same thing here. I would divide this entire expression by two. Which means I have to put a two in front of that expression to maintain the equivalents. Okay, so now what we would have is Ah too. One plus 35 to the n minus two. And then one plus five squared over too. Minus one plus Route five minus two equals two times one minus 15 to the N minus two. And then we have one minus five squared over two minus one minus with five minus two. And, of course, this, uh, expression is what we had in part be so we can use the simplified form. Now, this would be too one plus Route five to the N minus two and then three, plus route five minus one minus. Route five minus two equals two times one minus route five and minus two. And that would be three minus five minus one plus route five minus two. And, yeah, the front looks a little bit different, but the focus is going to be the big square brackets three minus one minus to cancel it to zero. Route five minus Route five cancels out to zero. So this effectively is just zero on the other side, three minus one minus two cancels out. Negative five Plus the five that also canceled out to zero. So, really, we have verified that the left side and the right side are both equal to zero because it's a multiplier and zero is equal to zero or the left side is equal to the right side. We have verified That's, um the expression works and we did it through using the result in part B.