154 Solutions and Hints for Selected Exercises SECTION 1.3 Exercise 1.3.5. For n = 1, √ √ √ 1 h1 + 5 1 − 5i 12 5 √ − = √ = 1. 5 2 2 5 2 Thus, the conclusion holds for n = 1. It is also easy to verify that the conclusion holds for n = 2. Suppose that √ √ 1 h 1 + 5 k 1 − 5 ki ak = √ − 5 2 2 for all k ≤ n, where n ≥ 2. Let us show that √ √ 1 h 1+ 5 n+1 1 − 5 n+1i an+1 = √ − . (5.12) 5 2 2 By the defnition of the sequence and the induction hypothesis, an+1 = an + an −1 √ √ √ √ h 1+ n 1 − ni h 1 + n−1 n−1i 1 5 5 1 5 1 − 5 = √ − + √ − 5 2 2 5 2 2 √ √ √ √ h 1 + i 1 5 n−1 1+ 5 1 − 5 n−1 1 − 5 = √ + 1 − + 1 . 5 2 2 2 2 Observe that √ √ √ √ √ √ 1 + 5 3 + 5 1+ 5 2 1 − 5 3− 5 1− 5 2 + 1 = = and + 1 = = . 2 2 2 2 2 2 Therefore, (5.12) follows easily. √ √ 1+ 5 1 − 5 In this exercise, observe that the two numbers and are the roots of the quadratic 2 2 equation 2x = x + 1. A more general result can be formulated as follows. Consider the sequence {an} defned by a1 = a; a2 = b; an+2 = αan+1 + β an for n ∈ N. 2 Suppose that the equation x = 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.
RkJQdWJsaXNoZXIy NTc4NTAz