00:01
This problem, we want to find the first 10 terms of this sequence.
00:05
The sequence that's given to us here, there's an expression, fn equals 1 over root 5, and then we got that complicated bracket.
00:15
So we're asked to find the first 10 terms and then compare them to the fibonacci number.
00:19
So let's find the first couple of terms and see if we can generate a pattern.
00:24
First term would be f of 1, f1.
00:26
That would be 1 over root 5, 1 plus root 5 to the power of 1, minus 1 minus root 5 to the power of 1 over 2 to the power of 1.
00:40
And we do have some like terms on the top.
00:45
1 minus 1 would cancel root 5 plus root 5 would give us 2 root 5 over 2, and that works out to 1.
00:56
F of 2 is done in a similar manner.
01:00
That would be 1 over root 5.
01:04
And then 1 plus root 5 squared minus 1 minus root 5 squared over 2 squared.
01:15
It equals 1 over root 5.
01:19
And we would expand that.
01:21
That would be 1 plus 2 root 5 plus 5 minus 1.
01:32
Minus 2 root 5 plus 5 all over 4 and on the top we can simplify we would have 1 plus 5 minus 5 so the constants go away 2 root 5 plus 2 root 5 would be 4 root 5 over 4 and that also works up to 1 f of 3 that would be 1 over root 5 and then we got 1 plus 1 plus 1 root 5 to the power of 3 minus 1 minus root 5 to the power of 3 all over 2 to the 3 which is 8 so yes we're going to have to expand some cubics there that's unfortunate so let's expand 1 plus root 5 over here on the right side 1 plus root 5 to the power of 3 is equal to 1 plus root 5 times 1 plus root 5 times 1 plus root 5 times 1 plus root 5 first two would combine to 1 plus 2 root 5 plus 5 and we can combine the like terms there's 6 plus 2 root 5 and now we can multiply it one more time that makes 6 plus 6 root 5 plus 2 root 5 plus 2 times 5 which is 10 so 16 plus 8 root 5 that's the expansion of the cubic and if we actually did one minus root 5 cube that actually works out to 16 minus 8 root 5 that's by a similar process so carrying those two answers forward here we would have 16 plus 8 root 5 minus 16 minus negative so plus 8 root 5 over 8 the 16s cancel we have 16 root 5 on top and 8 root 5 on the bottom which works out to 2 so we've got 1 1 and 2 you can continue to do f4 f5 f6 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 but if we had continued this process, it would be one, one, two, three, five, eight, 13, 21, 34, one, two, three, four, five, six, seven, eight, nine, and one more, 55.
04:34
Those are the first ten terms of fn using that formula.
04:41
Part b asks us to show that 3 plus or minus root 5 is equal to 1 plus or minus root 5 squared over.
05:01
So again, it's a show that problem.
05:04
We want to start with the left side and arrive at the answer on the right.
05:08
So it's a plus minus, so let's do it in two cases.
05:12
So let's do case one here.
05:14
Let's just work with 3 plus root 5.
05:17
So 3 plus root 5.
05:20
We want to make it look like a fraction, so we want that over 2 there.
05:24
That would be written as 6 plus 2 root 5 over 2.
05:31
And recognizing that we can turn that into a perfect square trinomial by saying 1 plus 2 root 5 plus 5 over 2, that turns it into 1 plus root 5 over 2.
05:45
2.
05:49
Case 2 would be starting with 3 minus root 5, the second case.
05:56
That would be similar.
05:58
That would be the same as 6 minus 2 root 5 over 2, which is the same as 1 plus, that 1 minus 2 root 5 over 2.
06:12
And that is a perfect square trinomial.
06:14
So that's 1 minus root 5 over 2.
06:18
So seeing that both cases work here, it just corresponds to the plus minus sign.
06:24
We can conclude that 3 plus or minus root 5 is equal to 1 plus or minus root 5 over 2, or squared, all over 2.
06:35
There's the answer to part b.
06:40
C, and this is the hard part.
06:43
It says use the result in part b to verify that fn satisfies the recursive definition of the fibonacci sequence.
06:51
So the recursive definition is fn equals the previous two terms.
06:57
So fn minus 1 plus fn minus 2.
07:04
And the problem, we're given that definition.
07:08
So we can replace the ends accordingly.
07:11
So this is the same as 1 over root 5 times 1 plus root 5 to the power of 1.
07:25
N minus 1 minus root 5 to the power of n all over 2 to the n equals 1 over root 5 and then 1 plus root 5 to the n minus 1 minus 1 minus root 5 to the n minus 1 all over 2 to the n minus 1 plus 1 over root 5 times 1 plus root 5 to the n minus 2 minus 1 minus root 5 to the n minus 2 all over 2 to the n minus 2.
08:19
So that's using the definition.
08:21
And from here, we can see that the 1 over root 5 is in common with all the terms...