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Math Help - Lucas Sequence

  1. #1
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    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!
    Last edited by MathStudent1; April 5th 2007 at 09:40 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    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 ?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    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.
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  4. #4
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    Quote Originally Posted by MathStudent1 View Post
    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.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
    Last edited by MathStudent1; April 5th 2007 at 11:18 AM.
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  6. #6
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    Quote Originally Posted by MathStudent1 View Post
    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.
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  7. #7
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    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!
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