17. The Lucas numbers are defined by the same recurrence formula as the Fibonacci numbers, Ln=Ln-1+ L-2 n > 3 but with L1 =1 and L2 = 3; this gives the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199,322, .... For the Lucas numbers, derive each of the identities below: (a) L1+ L2+ L3+ (b) L1+ L3 + L5+ (c) L2+ L4 + L6++L2n = L2n+1-1, n > 1. %3D + Ln = Ln+2-3, n > 1. +L2n-1 = L2n-2, n > 1. %D |
17. The Lucas numbers are defined by the same recurrence formula as the Fibonacci numbers, Ln=Ln-1+ L-2 n > 3 but with L1 =1 and L2 = 3; this gives the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199,322, .... For the Lucas numbers, derive each of the identities below: (a) L1+ L2+ L3+ (b) L1+ L3 + L5+ (c) L2+ L4 + L6++L2n = L2n+1-1, n > 1. %3D + Ln = Ln+2-3, n > 1. +L2n-1 = L2n-2, n > 1. %D |
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
14.3 question 17 part a,b, c please
![19. If a = and B = obtain the for the Lucas numbers
14. Verify that the productu-u1287e4 of three consecutive Fibonacci numbers with even
indices is the product of three consecutive integers; for instance, we have u4ugu8 =
504 7-8.9.
[Hint: First show that uzu2n+4=u42-1.]
15. Use Eqs. (1) and (2) to show that the sum of any 20 consecutive Fibonacci numbers is
divisible by u 10-
16. For n 24, prove that un +1 is not a prime.
[Hint: It suffices to establish the identities
%3D
(2+n + n)I-Kn = [+ n
(I+7n +1-Rn)I+Tn = [ + +n
(1-Rn + I+tn)Z+Tn = [ +
['(E+Tn + I+rn)I+Tn= [ + E+n
17. The Lucas numbers are defined by the same recurrence formula as the Fibonacci numbers,
La= Ln-1+ L-2
n> 3
but with L1 =1 and L2 = 3; this gives the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 123,
For the Lucas numbers, derive each of the identities below:
+ Ln = Ln+2 - 3, n > 1.
+L2n-1 = L2n-2, n > 1.
+L2n = L2n+1-1, n > 1.
%3D
199, 322,
(a) L1+ L2 + L3 +
(b) L1+ L3 + L5+
(c) L2+ L4+ L6+
(d) L= Ln+1Ln-1+5(-1)", n > 2.
(e) L+ L+L++L L„Ln+1 - 2, n > 1.
(f)L-L= L-1Ln+2, n 2 2.
18. Establish the following relations between the Fibonacci and Lucas numbers:
(a) Ln = un+1 +un-1 = un + 2un-1, n > 2.
[Hint: Argue by induction on n.]
%3D
%3D
|
%3D
%3D
%3D
%3D
(c) u2n =unLn, n > 1.
(d)Ln+1+Ln-1 = 5un, n 2 2.
(e)3u2+4un+14n-1, n > 2.
(1) 2um+n =umL, + Lmun, m 2 1, n 2 1.
() gcd(un, Ln) =1 or 2, n 1.
%3D
5)/2 and B = (1 - 5/2. obtain the Binet formula for the Lucas numbers
n > 1
%3D
Ln =a"+ B"
(c) L-L-1Lnt
- 5(-1 n > 2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c7b497-dac7-4855-b923-2e60bbc73063%2Ff9b1c469-287c-4df1-90bc-ff83209b3812%2Fy6zq8uj.jpeg&w=3840&q=75)
Transcribed Image Text:19. If a = and B = obtain the for the Lucas numbers
14. Verify that the productu-u1287e4 of three consecutive Fibonacci numbers with even
indices is the product of three consecutive integers; for instance, we have u4ugu8 =
504 7-8.9.
[Hint: First show that uzu2n+4=u42-1.]
15. Use Eqs. (1) and (2) to show that the sum of any 20 consecutive Fibonacci numbers is
divisible by u 10-
16. For n 24, prove that un +1 is not a prime.
[Hint: It suffices to establish the identities
%3D
(2+n + n)I-Kn = [+ n
(I+7n +1-Rn)I+Tn = [ + +n
(1-Rn + I+tn)Z+Tn = [ +
['(E+Tn + I+rn)I+Tn= [ + E+n
17. The Lucas numbers are defined by the same recurrence formula as the Fibonacci numbers,
La= Ln-1+ L-2
n> 3
but with L1 =1 and L2 = 3; this gives the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 123,
For the Lucas numbers, derive each of the identities below:
+ Ln = Ln+2 - 3, n > 1.
+L2n-1 = L2n-2, n > 1.
+L2n = L2n+1-1, n > 1.
%3D
199, 322,
(a) L1+ L2 + L3 +
(b) L1+ L3 + L5+
(c) L2+ L4+ L6+
(d) L= Ln+1Ln-1+5(-1)", n > 2.
(e) L+ L+L++L L„Ln+1 - 2, n > 1.
(f)L-L= L-1Ln+2, n 2 2.
18. Establish the following relations between the Fibonacci and Lucas numbers:
(a) Ln = un+1 +un-1 = un + 2un-1, n > 2.
[Hint: Argue by induction on n.]
%3D
%3D
|
%3D
%3D
%3D
%3D
(c) u2n =unLn, n > 1.
(d)Ln+1+Ln-1 = 5un, n 2 2.
(e)3u2+4un+14n-1, n > 2.
(1) 2um+n =umL, + Lmun, m 2 1, n 2 1.
() gcd(un, Ln) =1 or 2, n 1.
%3D
5)/2 and B = (1 - 5/2. obtain the Binet formula for the Lucas numbers
n > 1
%3D
Ln =a"+ B"
(c) L-L-1Lnt
- 5(-1 n > 2.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 3 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
