The Lucas numbers L(n) have almost the same definition as the Fibonacci numbers: if n = 1 if n = 2 L(n-1)+L(n − 2) if n > 2. 5 as in Theorem 3.6. Prove that L(n) = a + ßn for all n E N. Use strong induction. L(n): Let a = = = 1+√5 and ß = 2 1 3 and Proof. First, note that 1 - L(1) = 1 = a + ß, a² + ² = (a + 1) + ( + 2 = a + B + 2 = L(2). Suppose as inductive hypothesis that L(i) = a¹ + ßi for all i 2. Then +L(K- = L(K) = L(k − 1) + = ak-1 + Bk - 1 + = ak − 2(a + 1) + ßk − 2 (B+ ak-2(a²) + pk-2( +

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