Discussion
The Lucas sequence \(L_{n}\) is defined by the recurrence relation:
\[ L_{n} = L_{n-1} + L_{n-2}, \quad \text{for} \quad n \ge 3, \]
with \(L_{1} = 1\) and \(L_{2} = 3\).
Which one of the options given is TRUE?
Answer: A
The true formula is \(L_{n} = (\frac{1+\sqrt{5}}{2})^{n} + (\frac{1-\sqrt{5}}{2})^{n}\).
Write \(\varphi = \frac{1+\sqrt{5}}{2}\) and \(\psi = \frac{1-\sqrt{5}}{2}\), the roots of \(x^{2} = x + 1\). Any combination \(a\varphi^{n} + b\psi^{n}\) satisfies the recurrence, so it only remains to match the first two terms. With \(\varphi + \psi = 1\) and \(\varphi\psi = -1\):
\[ \begin{aligned} \varphi + \psi &= 1 = L_{1} \\ \varphi^{2} + \psi^{2} &= (\varphi + \psi)^{2} - 2\varphi\psi = 3 = L_{2} \end{aligned} \]
The difference \(\varphi^{n} - \psi^{n}\) equals \(\sqrt{5}\) at n = 1, not 1; it is \(\sqrt{5}\) times the Fibonacci numbers. The two options with denominator 3 use a number that is not a root of \(x^{2} = x + 1\), so they do not even obey the recurrence; at n = 1 the '+' version gives about 1.21.
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