1. ## Lucas Sequence

Here is my problem, two parts.

The Lucas sequence is defined by

(L_n) = {1,3,4,7,11,18,...}

where the successive terms is found by adding the two previous terms.

1.) Show that the general term, or the nth term of the Lucas sequence is

L_n = [(1 + SQRT(5))/2]^n + [(1 - SQRT(5))/2]^n = (alpha)^n + (beta)^n

2.) Prove the formula for the general term using mathematical induction.

***********

Thank You!

2. are you sure there's not an n somewhere in L_n = [(1 + SQRT(5))/2]^2 + [(1 - SQRT(5))/2]^2 = (alpha)^2 + (beta)^2 ?

3. Originally Posted by Jhevon
are you sure there's not an n somewhere in L_n = [(1 + SQRT(5))/2]^2 + [(1 - SQRT(5))/2]^2 = (alpha)^2 + (beta)^2 ?
Yes, Thank you Jhevon I have too many twos. I corrected it.

4. Originally Posted by MathStudent1
Here is my problem, two parts.

The Lucas sequence is defined by

(L_n) = {1,3,4,7,11,18,...}

where the successive terms is found by adding the two previous terms.

1.) Show that the general term, or the nth term of the Lucas sequence is

L_n = [(1 + SQRT(5))/2]^n + [(1 - SQRT(5))/2]^n = (alpha)^n + (beta)^n
The reccurence relation is,

L_n = L_(n-1)+L_(n-2)

Charachteristic polynomial is,

x^2 = x + 1

The solutions are,

x=(1+sqrt(5))/2, (1-sqrt(5))/2

The general solution is,

L_n = A[(1+sqrt(5))/2]^n+B[(1-sqrt(5))/2]^n

The initial conditios are,

L_1=1 and L_2=2

Use this to determine A and B.
And that leads to this formula.

5. Originally Posted by ThePerfectHacker
The reccurence relation is,

L_n = L_(n-1)+L_(n-2)

Charachteristic polynomial is,

x^2 = x + 1

The solutions are,

x=(1+sqrt(5))/2, (1-sqrt(5))/2

The general solution is,

L_n = A[(1+sqrt(5))/2]^n+B[(1-sqrt(5))/2]^n

The initial conditios are,

L_1=1 and L_2=2

Use this to determine A and B.
And that leads to this formula.
Question, how do you determine your initial conditions? I thought it would be L_0 = 1 and L_1 = 3? I understand everything to that point.

6. Originally Posted by MathStudent1
Question, how do you determine your initial conditions? I thought it would be L_0 = 1 and L_1 = 3? I understand everything to that point.
It depends whether you accept that the zero-th term is actually the first. If thus,

L_0 = 1 and L_1 =3

Yes, I made a mistake. I accidently use the Fibonacci sequence.

7. Okay, I am stuck in the algebra. Here is what I have.

(L_n) = {1,3,4,7,11,18,...}

and is defined recurrsively as follows:

a_0 = 1
a_1 = 3
a_n = a_n-1 + a_n-2

a_n - a_n-1 - a_n-2 = 0
x^2 - x - 1 = 0

For ax^2 + bx + c = 0

x = -b (+,-) SQRT(b^2 - 4ac) / 2a

x = 1 (+,-) SQRT(1 - 4(-1)) / 2

x = 1 (+,-) SQRT(1 (+,-) SQRT(5) / 2

Then, a_n = A[(1+sqrt(5))/2]^n+B[(1-sqrt(5))/2]^n

Must find A and B

Let n = 0

a_0 = A + B = 1

Let n = 1

a_1 = A[(1+sqrt(5))/2]^1+B[(1-sqrt(5))/2]^1 = 3

C = [[1,1,1][(1+sqrt(5))/2,(1-sqrt(5))/2,3]] (matrix)

rref(A) gives A = (1+sqrt(5))/2 and B = (1-sqrt(5))/2

substituting in A and B gives

a_n = ((1+sqrt(5))/2)[(1+sqrt(5))/2]^n + ((1-sqrt(5))/2)[(1-sqrt(5))/2]^n

a_n = [((1+sqrt(5))/2]^(n+1) + [((1-sqrt(5))/2]^(n+1)

Let (alpha) = ((1+sqrt(5))/2 and (beta) = ((1-sqrt(5))/2

Then, L_n = (alpha)^(n+1) + (beta)^(n+1) general term

*****

b.) Prove P_n : a_n = [((1+sqrt(5))/2]^(n+1) + [((1-sqrt(5))/2]^(n+1)

Step 1: Using initial conditions.

Let n = 0

a_0 = [((1+sqrt(5))/2]^1 + [((1-sqrt(5))/2]^1 = 1

a_0 = 1 = 1 it checks!

Let n = 1

a_1 = [((1+sqrt(5))/2]^2 + [((1-sqrt(5))/2]^1 = 3

a_1 = 12/4 = 3 it checks!

Step 2:

Assume a_n-1 and a_n-2 are both true

a_n-1 = [((1+sqrt(5))/2]^(n-1) + [((1-sqrt(5))/2]^(n-1)

a_n-2 = [((1+sqrt(5))/2]^(n-2) + [((1-sqrt(5))/2]^(n-2)

Step 3:

Must show true for a_n. That is to say must show:

a_n = [((1+sqrt(5))/2]^(n+1) + [((1-sqrt(5))/2]^(n+1)

*using recurrence relations gives

a_n = a _n-1 + a_n-2

Let (alpha) = [((1+sqrt(5))/2]

Let (beta) = [((1-sqrt(5))/2]

Now,

a_n = [(alpha)^(n-1) + (beta)^(n-1)] + [(alpha)^(n-2) + (beta)^(n-2)]

OKAY, FROM HERE I AM STUCK. I CERTAINLY COULD OF MADE A MISTAKE ABOVE!