Page 311 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Strong Induction 297
and

2n − 1 2n − 2 2n − 3 n
F 2n = + + +· · · +
0 1 2 n − 1
n−1
2n − i − 1
= .
i
i=0
∗ 8. A sequence of numbers a 0 , a 1 , a 2 , ... is called a generalized Fibonacci
sequence,ora Gibonacci sequence for short, if for every n ≥ 2,
a n = a n−2 + a n−1 . Thus, a Gibonacci sequence satisfies the same recur-
rence relation as the Fibonacci numbers, but it may start out differently.
n
(a) Suppose c is a real number and ∀n ∈ N(a n = c ). Prove that a 0 ,
√
a 1 , a 2 , ... is a Gibonacci sequence iff either c = (1 + 5)/2or
√
c = (1 − 5)/2.
(b) Suppose s and t are real numbers, and for all n ∈ N,
n n
√ √
1 + 5 1 − 5
a n = s + t .
2 2
Prove that a 0 , a 1 , a 2 , ... is a Gibonacci sequence.
(c) Suppose a 0 , a 1 , a 2 , ... is a Gibonacci sequence. Prove that there are
real numbers s and t such that for all n ∈ N,
n n
√ √

1 + 5 1 − 5
a n = s + t .
2 2
(Hint: First show that there are real numbers s and t such that the
formula above is correct for a 0 and a 1 . Then show that with this
choice of s and t, the formula is correct for all n.)
9. The Lucas numbers (named for the French mathematician Edouard Lucas
(1842–1891)) are the numbers L 0 , L 1 , L 2 , ... defined as follows:

L 0 = 2;
L 1 = 1;
for every n ≥ 2, L n = L n−2 + L n−1 .
Find a formula for L n and prove that your formula is correct. (Hint: Apply
exercise 8.)
10. A sequence a 0 , a 1 , a 2 , ... is defined recursively as follows:
∗
a 0 =−1;
a 1 = 0;
for every n ≥ 2, a n = 5a n−1 − 6a n−2 .