Introduction to Mathematical Analysis I - Second Edition

144 Solutions and Hints for Selected Exercises SECTION 1.3 Exercise 1.3.6 . For n = 1, 1 √ 5 h 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 a k = 1 √ 5 h 1 + √ 5 2 k − 1 − √ 5 2 k i for all k ≤ n , where n ≥ 2. Let us show that a n + 1 = 1 √ 5 h 1 + √ 5 2 n + 1 − 1 − √ 5 2 n + 1 i . (5.1) By the definition of the sequence and the induction hypothesis, a n + 1 = a n + a n − 1 = 1 √ 5 h 1 + √ 5 2 n − 1 − √ 5 2 n i + 1 √ 5 h 1 + √ 5 2 n − 1 − 1 − √ 5 2 n − 1 i = 1 √ 5 h 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, ( 5.1 ) follows easily. In this exercise, observe that the two numbers 1 + √ 5 2 and 1 − √ 5 2 are the roots of the quadratic equation x 2 = x + 1 . A more general result can be formulated as follows. Consider the sequence { a n } defined by a 1 = a ; a 2 = b ; a n + 2 = α a n + 1 + β a n for n ∈ N . Suppose that the equation x 2 = α x + β has two solutions x 1 and x 2 . Let c 1 and c 2 be two constants such that c 1 x 1 + c 2 x 2 = a ; c 1 ( x 1 ) 2 + c 2 ( x 2 ) 2 = b . Then we can prove by induction that x n = c 1 ( x 1 ) n + c 2 ( x 2 ) n for all n ∈ N .