Use mathematical induction to prove that for each integer n2 0, if F, F F ... is the Fibonacci sequence, then ged(F.. Fa) - 1. (The definition of ged is given in Section 4.10.) You may use the following lemma in the proof. Lemma: If a and b are any integers not both zero, and if g and rare any integers such that a- bg +r, then god(a, b) - ged(b, r). Proof (by mathematical induction): Let the property P(n) be the equation ged(F. F) - 1. Show that P(0) is true: Select P(0) from the choices below. OF, F1 O gcd(F F) = 1 • ged(F,. F) = 1 O gcd(F,. F;) = 1 The selected statement is true by definition of the Fibonacci sequence. Show that for each integer k2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with kz 0, and suppose that P(k) is true. Select P(k) from the choices below. • gcd(F.F) -1 O gcd(F, . 3 Fa + 3) = 1 O ged(F F) - 1 O gcd(F, . F+ ) = 1 (This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. Select P(k + 1) from the choices below. O ged(Fz F) - 1 • gcd(F,. F.)-1 O ged(F,. 3 Fn ) = 1 O gcd(F, . 3 F,) = 1 Now F F..+F by defntion of the Fibonacci sequence V . Moreover, since F,.2=F+F, can be rewritten as F,+2F+1+F, the lemma can be applied with a= and r Select the result of applying the lemma from the choices below. O ged(F,. F) - ged(F,. F) O gcd(F, - 1 F) = gcd(F, - 1. F;) • ged(F.. 2. F. ;) = gcd(F, . 1• F2) So, by applying the inductive hypothesis, we can conclude that ged(F [as was to be shown].
Use mathematical induction to prove that for each integer n2 0, if F, F F ... is the Fibonacci sequence, then ged(F.. Fa) - 1. (The definition of ged is given in Section 4.10.) You may use the following lemma in the proof. Lemma: If a and b are any integers not both zero, and if g and rare any integers such that a- bg +r, then god(a, b) - ged(b, r). Proof (by mathematical induction): Let the property P(n) be the equation ged(F. F) - 1. Show that P(0) is true: Select P(0) from the choices below. OF, F1 O gcd(F F) = 1 • ged(F,. F) = 1 O gcd(F,. F;) = 1 The selected statement is true by definition of the Fibonacci sequence. Show that for each integer k2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with kz 0, and suppose that P(k) is true. Select P(k) from the choices below. • gcd(F.F) -1 O gcd(F, . 3 Fa + 3) = 1 O ged(F F) - 1 O gcd(F, . F+ ) = 1 (This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. Select P(k + 1) from the choices below. O ged(Fz F) - 1 • gcd(F,. F.)-1 O ged(F,. 3 Fn ) = 1 O gcd(F, . 3 F,) = 1 Now F F..+F by defntion of the Fibonacci sequence V . Moreover, since F,.2=F+F, can be rewritten as F,+2F+1+F, the lemma can be applied with a= and r Select the result of applying the lemma from the choices below. O ged(F,. F) - ged(F,. F) O gcd(F, - 1 F) = gcd(F, - 1. F;) • ged(F.. 2. F. ;) = gcd(F, . 1• F2) So, by applying the inductive hypothesis, we can conclude that ged(F [as was to be shown].
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Use mathematical induction to prove that for each integer \( n \geq 0 \), if \( F_0, F_1, F_2, \ldots \) is the Fibonacci sequence, then
\[ \text{gcd}(F_{n+1}, F_n) = 1. \]
(The definition of gcd is given in Section 4.1.0.)
You may use the following lemma in the proof.
**Lemma**: If \( a \) and \( b \) are any integers not both zero, and if \( q \) and \( r \) are any integers such that \( a = bq + r \), then \( \text{gcd}(a, b) = \text{gcd}(b, r) \).
**Proof (by mathematical induction)**: Let the property \( P(n) \) be the equation \( \text{gcd}(F_{n+1}, F_n) = 1 \).
**Show that \( P(0) \) is true**: Select \( P(0) \) from the choices below.
- \( F_1 = F_0 = 1 \)
- \( \text{gcd}(F_1, F_0) = 0 \)
- \( \text{gcd}(F_1, F_0) = 1 \) (selected)
- \( \text{gcd}(F_2, F_1) = 1 \)
The selected statement is true by definition of the Fibonacci sequence.
**Show that for each integer \( k \geq 0 \), if \( P(k) \) is true, then \( P(k + 1) \) is true**: Let \( k \) be any integer with \( k \geq 0 \), and suppose that \( P(k) \) is true. Select \( P(k) \) from the choices below.
- \( \text{gcd}(F_{k+1}, F_k) = 0 \)
- \( \text{gcd}(F_{k+1}, F_k) = 1 \) (selected)
- \( \text{gcd}(F_{k+2}, F_{k+1}) = 1 \)
- \( \text{gcd}(F_{n+2} + F_{n+1}, 1) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2b801ef-259c-473d-9009-1e1b33ddc8b5%2F835145bc-87ba-4b6b-8ddf-88089214806a%2Fzbywgd_processed.png&w=3840&q=75)
Transcribed Image Text:Use mathematical induction to prove that for each integer \( n \geq 0 \), if \( F_0, F_1, F_2, \ldots \) is the Fibonacci sequence, then
\[ \text{gcd}(F_{n+1}, F_n) = 1. \]
(The definition of gcd is given in Section 4.1.0.)
You may use the following lemma in the proof.
**Lemma**: If \( a \) and \( b \) are any integers not both zero, and if \( q \) and \( r \) are any integers such that \( a = bq + r \), then \( \text{gcd}(a, b) = \text{gcd}(b, r) \).
**Proof (by mathematical induction)**: Let the property \( P(n) \) be the equation \( \text{gcd}(F_{n+1}, F_n) = 1 \).
**Show that \( P(0) \) is true**: Select \( P(0) \) from the choices below.
- \( F_1 = F_0 = 1 \)
- \( \text{gcd}(F_1, F_0) = 0 \)
- \( \text{gcd}(F_1, F_0) = 1 \) (selected)
- \( \text{gcd}(F_2, F_1) = 1 \)
The selected statement is true by definition of the Fibonacci sequence.
**Show that for each integer \( k \geq 0 \), if \( P(k) \) is true, then \( P(k + 1) \) is true**: Let \( k \) be any integer with \( k \geq 0 \), and suppose that \( P(k) \) is true. Select \( P(k) \) from the choices below.
- \( \text{gcd}(F_{k+1}, F_k) = 0 \)
- \( \text{gcd}(F_{k+1}, F_k) = 1 \) (selected)
- \( \text{gcd}(F_{k+2}, F_{k+1}) = 1 \)
- \( \text{gcd}(F_{n+2} + F_{n+1}, 1) =
![### Recursive Sequence Definition and Iteration
Consider the following recursively defined sequence:
\[
f_k = f_{k-1} + 2^k, \text{ for each integer } k \geq 2
\]
\[
f_1 = 1
\]
---
### Iteration to Guess an Explicit Formula
Fill in the blanks to use iteration to guess an explicit formula for the sequence.
1. \( f_1 = 1 \)
2. \( f_2 = f_1 + \boxed{2^2} = 1 + 2^2 \)
3. \( f_3 = f_2 + \boxed{2^3} = 1 + 2^2 + 2^3 \)
4. \( f_4 = f_3 + \boxed{2^4} = 1 + 2^2 + 2^3 + 2^4 \)
---
**Guess:**
\[
f_n = 1 + 2^2 + 2^3 + 2^4 + \ldots + \boxed{2^n}
\]
---
### Simplification Using Theorem 5.2.2
When Theorem 5.2.2 is used to simplify this expression, the result is:
\[
f_n = \left( \frac{\boxed{2^{n+1}} - 1}{2 - 1} \right) - 2
\]
And, when this expression is simplified, the result is:
\[
f_n = \boxed{2^{n+1}} - 3 \text{ for every integer } n \geq 1.
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2b801ef-259c-473d-9009-1e1b33ddc8b5%2F835145bc-87ba-4b6b-8ddf-88089214806a%2Fvkzcuch_processed.png&w=3840&q=75)
Transcribed Image Text:### Recursive Sequence Definition and Iteration
Consider the following recursively defined sequence:
\[
f_k = f_{k-1} + 2^k, \text{ for each integer } k \geq 2
\]
\[
f_1 = 1
\]
---
### Iteration to Guess an Explicit Formula
Fill in the blanks to use iteration to guess an explicit formula for the sequence.
1. \( f_1 = 1 \)
2. \( f_2 = f_1 + \boxed{2^2} = 1 + 2^2 \)
3. \( f_3 = f_2 + \boxed{2^3} = 1 + 2^2 + 2^3 \)
4. \( f_4 = f_3 + \boxed{2^4} = 1 + 2^2 + 2^3 + 2^4 \)
---
**Guess:**
\[
f_n = 1 + 2^2 + 2^3 + 2^4 + \ldots + \boxed{2^n}
\]
---
### Simplification Using Theorem 5.2.2
When Theorem 5.2.2 is used to simplify this expression, the result is:
\[
f_n = \left( \frac{\boxed{2^{n+1}} - 1}{2 - 1} \right) - 2
\]
And, when this expression is simplified, the result is:
\[
f_n = \boxed{2^{n+1}} - 3 \text{ for every integer } n \geq 1.
\]
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