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 ?
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.
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Thank You!
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.
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
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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!