00:02
We are given the definition of the lucas numbers and initial conditions for the lucas numbers.
00:09
We are asked to answer some questions about them.
00:12
So they satisfy the recurrence relation ln equals ln minus 1 plus ln minus 2, and the initial conditions l0 equals 2 and l1 equals 1.
00:25
In part a, we're asked to write the lucas number ln in terms of the nth, in terms of the n minus 1st and n plus first, fibonacci numbers.
00:48
So in this case, we'll use induction.
00:55
So if n is equal to 2, we have that l2, this is equal to from the definition of lucas numbers, l1 plus l0, which is 3, and we have that f1 plus f3, this is going to be 1 plus 2, which is also 3, so that it works for the base case and for the inductive step we're going to suppose that the lucas number ln is equal to fn minus 1 plus fn plus 1 for all n less than or equal to k for some k which is going to be greater than or equal to 2 and this is going to be all n between 2 and k so we have that l sub k plus 1 by the recurrence relation is lk plus lk minus 1.
02:54
By the inductive hypothesis, this is equal to fk minus 1 plus fk plus 1, which is equal to or plus and this is fk minus 2 plus fk.
03:23
And this can be rewritten as fk minus 2 plus fk minus 1 plus fk plus fk plus 1 and we have by the recurrence relation for the fibinacci numbers that this is equal to fk plus fk plus 2 and this is what we wanted to show so it follows that l n is equal to fn minus 1 plus fn plus 1 for all natural numbers greater than are equal to 2 by induction.
04:26
And in part b, we're asked to find an explicit formula for the lucas numbers.
04:37
So we have the lucas numbers have a linear homogenous recurrence relation.
04:42
And so the characteristic equation is r squared minus r minus 1 equals 0, which has roots r equals 1 plus or minus the square root of 1 plus 4, which is 5, all over 2...