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( +

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The Lucas numbers L(n) have almost the same definition as the Fibonacci numbers:
1
1
if n =
if n = 2
3
L(n − 1)+L(n − 2) if n > 2.
Let a =
L(n) =
=
1 + 5
2
and ß:
=
and
Proof. First, note that
L(1) = 1 = a + ß,
1 - √5
2
a² + B² = (a + 1) + (B +
=
= L(2).
Suppose as inductive hypothesis that L(i)
)+L(K=
=
L(K) L(k − 1) + L
=
= a + ß + 2
=
= ak - 1
=
I
ak-2(a +
as in Theorem 3.6. Prove that L(n) = a + ßn for all n E N. Use strong induction.
ak +
+ Bk - 1
+ 1)
+
+
= ak − ²(a²) + ßk -
=
BK-2 (B.
+
ai + Bi for all i<k, for some k > 2. Then
Transcribed Image Text:The Lucas numbers L(n) have almost the same definition as the Fibonacci numbers: 1 1 if n = if n = 2 3 L(n − 1)+L(n − 2) if n > 2. Let a = L(n) = = 1 + 5 2 and ß: = and Proof. First, note that L(1) = 1 = a + ß, 1 - √5 2 a² + B² = (a + 1) + (B + = = L(2). Suppose as inductive hypothesis that L(i) )+L(K= = L(K) L(k − 1) + L = = a + ß + 2 = = ak - 1 = I ak-2(a + as in Theorem 3.6. Prove that L(n) = a + ßn for all n E N. Use strong induction. ak + + Bk - 1 + 1) + + = ak − ²(a²) + ßk - = BK-2 (B. + ai + Bi for all i<k, for some k > 2. Then
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