n+1 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 onn.] - un-2, n > 3. %3D %3D Иn-2, п > 3. n > 1 (b) Ln = un+2 %3D
n+1 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 onn.] - un-2, n > 3. %3D %3D Иn-2, п > 3. n > 1 (b) Ln = un+2 %3D
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
18 part a, b thanks
![=7 (mod 10) for n > 2
) L2n+1 5u,un41+(-1)", n z 1.
(0) L-나
2Lm+n, m 2 1, n > 1.
(d) Lm Ln +5umUn
%3D
all have 7 as the final digit;
%3D
%3D
%3D
%3D
%3D
5. Ita = (1+ 5)/2 and B = (1- V5)/2, obtain the Binet formula for the Lucas num
(g) gcd(un, Ln) =1 or 2, n > 1.
(f) 2um +n = um Ln+ LmUn, m > 1, n > 1.
(e) L u +4n+1Un-1, n > 2.
(d) Ln+1 + Ln-1 = 5un, n > 2.
(c) u2n =Un Ln, n> 1.
%3D
Ln =a"+ B"
%3D
n>1
%3D
%3D
%3D
%3D
%3D
un = "7 (9)
[Hint: Argue by induction on n.]
(a) Ln=un+1 + un-1 =un + 2un-1, n > 2.
18. Establish the following relations between the Fibonacci and Lucas numbers:
(f) L1-L = Ln-1Ln+2, n > 2.
(e) L구 + L3+ L3+ + L= Ln Ln+1 - 2, n > 1.
(d) L2 = Ln+1Ln-1+5(-1)", n > 2.
() L2 + L4 + L6+ + L2n = L2n+1-1, n > 1.
) L+ L3 + Ls + +
(a) Li+ L2 + L3 + +Ln = Ln+2-3, n > 1.
109 322, .... For the Lucas numbers, derive each of the identities below:
but with L1 =1 and L2 = 3; this gives the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 12
%3D
%3D
+ L = LnLn+1 - 2, n > 1.
%3D
%3D
+ L2n-1 = L2n - 2, n > 1.
%3D
%3D
The Lucas numbers are defined by the same recurrence formula as the Fibonacci numbe
%3D
['(E+n+I+n)I+n= [ +
%3D
u4k+2+13u2+2(u2k+1 + u2k-1)
2k+1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c7b497-dac7-4855-b923-2e60bbc73063%2F9ad5e6fe-5f6e-4922-968d-7d2c1ed75def%2Fckbbbwf.jpeg&w=3840&q=75)
Transcribed Image Text:=7 (mod 10) for n > 2
) L2n+1 5u,un41+(-1)", n z 1.
(0) L-나
2Lm+n, m 2 1, n > 1.
(d) Lm Ln +5umUn
%3D
all have 7 as the final digit;
%3D
%3D
%3D
%3D
%3D
5. Ita = (1+ 5)/2 and B = (1- V5)/2, obtain the Binet formula for the Lucas num
(g) gcd(un, Ln) =1 or 2, n > 1.
(f) 2um +n = um Ln+ LmUn, m > 1, n > 1.
(e) L u +4n+1Un-1, n > 2.
(d) Ln+1 + Ln-1 = 5un, n > 2.
(c) u2n =Un Ln, n> 1.
%3D
Ln =a"+ B"
%3D
n>1
%3D
%3D
%3D
%3D
%3D
un = "7 (9)
[Hint: Argue by induction on n.]
(a) Ln=un+1 + un-1 =un + 2un-1, n > 2.
18. Establish the following relations between the Fibonacci and Lucas numbers:
(f) L1-L = Ln-1Ln+2, n > 2.
(e) L구 + L3+ L3+ + L= Ln Ln+1 - 2, n > 1.
(d) L2 = Ln+1Ln-1+5(-1)", n > 2.
() L2 + L4 + L6+ + L2n = L2n+1-1, n > 1.
) L+ L3 + Ls + +
(a) Li+ L2 + L3 + +Ln = Ln+2-3, n > 1.
109 322, .... For the Lucas numbers, derive each of the identities below:
but with L1 =1 and L2 = 3; this gives the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 12
%3D
%3D
+ L = LnLn+1 - 2, n > 1.
%3D
%3D
+ L2n-1 = L2n - 2, n > 1.
%3D
%3D
The Lucas numbers are defined by the same recurrence formula as the Fibonacci numbe
%3D
['(E+n+I+n)I+n= [ +
%3D
u4k+2+13u2+2(u2k+1 + u2k-1)
2k+1)
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