154 Solutions and Hints for Selected Exercises SECTION 1.3 Exercise 1.3.5. For n=1, 1 √5h 1+√5 2 − 1− √5 2 i = 1 √5 2√5 2 =1. Thus, the conclusion holds for n=1. It is also easy to verify that the conclusion holds for n=2. Suppose that ak = 1 √5h 1+√5 2 k − 1− √5 2 ki for all k ≤n, where n≥2. Let us show that an+1 = 1 √5h 1+√5 2 n+1 − 1− √5 2 n+1i. (5.12) By the definition of the sequence and the induction hypothesis, an+1 =an +an−1 = 1 √5h 1+√5 2 n − 1− √5 2 ni + 1 √5h 1+√5 2 n−1 − 1− √5 2 n−1i = 1 √5h 1+√5 2 n−1 1+√5 2 +1 − 1− √5 2 n−1 1− √5 2 +1 i. Observe that 1+√5 2 +1= 3+√5 2 = 1+√5 2 2 and 1− √5 2 +1= 3− √5 2 = 1− √5 2 2 . Therefore, (5.12) follows easily. In this exercise, observe that the two numbers 1+√5 2 and 1− √5 2 are the roots of the quadratic equation x2 =x+1. A more general result can be formulated as follows. Consider the sequence {an}defined by a1 =a; a2 =b; an+2 =αan+1 +βan for n∈N. Suppose that the equation x2 =αx+β has two solutions x 1 and x2. Let c1 and c2 be two constants such that c1x1 +c2x2 =a; c1(x1) 2 +c 2(x2) 2 =b. Then we can prove by induction that xn =c1(x1) n +c 2(x2) n for all n∈N.
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