SOLUTION: The Fibonacci sequence, is defined by F_0 = 0, F_1 = 1, and F_n = F_{n - 2} + F_{n - 1}. It turns out that F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}, where \alpha = \frac{1 + \s

Question 1209828: The Fibonacci sequence, is defined by F_0 = 0, F_1 = 1, and F_n = F_{n - 2} + F_{n - 1}. It turns out that
F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}},
where \alpha = \frac{1 + \sqrt{5}}{2} and \beta = \frac{1 - \sqrt{5}}{2}.

The Lucas sequence is defined as follows: L_0 = 2, L_1 = 1, and
L_n = L_{n - 1} + L_{n - 2}
for n \ge 2. What is L_4?

Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's calculate $L_4$ using the given recurrence relation and initial values.
* $L_0 = 2$
* $L_1 = 1$
* $L_n = L_{n-1} + L_{n-2}$ for $n \ge 2$
Now, let's find the subsequent terms:
* $L_2 = L_1 + L_0 = 1 + 2 = 3$
* $L_3 = L_2 + L_1 = 3 + 1 = 4$
* $L_4 = L_3 + L_2 = 4 + 3 = 7$
Therefore, $L_4 = 7$.
Final Answer: The final answer is $\boxed{7}$

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The sequence with first two terms 2 and 1 with the recursive definition that each term is the sum of the previous two terms is NOT "THE" Lucas sequence. A Lucas sequence is ANY sequence in which each term is a linear combination of the two preceding terms (and the first two terms can be any numbers).

The sequence in this problem is A Lucas sequence with first two terms 2 and 1.

Subsequent terms of the sequence are found using the given recursive definition.

L(0)=2
L(1)=1
L(2)=L(0)+L(1)=2+1=3
L(3)=L(1)+L(2)=1+3=4
L(4)=L(2)+L(3)=3+4=7
L(5)=L(3)+L(4)+4+7=11
etc...

ANSWER: L(4)=7

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