Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation k = fr-1+ 2k for each integer k 2 2 f, = 1 satisfies the following formula. f. = 2n +1- 3 for every integer n 2 1 Proof (by mathematical induction): Suppose f,, f2, fa, ... is a sequence that satisfies the recurrence relation f = fk -1+ 2k for each integer k 2 2, with initial condition f, = 1. We need to show that when the sequence f, fa, far :.. is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above. So let the property P(n) be the equation f, = 2" +1- 3. We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: The left-hand side of P(1) is , which equals The right-hand side of P(1) is Since the left-hand and right-hand sides equal each other, P(1) is true. Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = , where f,, f,,f... is a sequence defined by the recurrence relation f, = fr -1+ 2k for each integer k 2 2, with initial condition f, = 1.
Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation k = fr-1+ 2k for each integer k 2 2 f, = 1 satisfies the following formula. f. = 2n +1- 3 for every integer n 2 1 Proof (by mathematical induction): Suppose f,, f2, fa, ... is a sequence that satisfies the recurrence relation f = fk -1+ 2k for each integer k 2 2, with initial condition f, = 1. We need to show that when the sequence f, fa, far :.. is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above. So let the property P(n) be the equation f, = 2" +1- 3. We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: The left-hand side of P(1) is , which equals The right-hand side of P(1) is Since the left-hand and right-hand sides equal each other, P(1) is true. Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = , where f,, f,,f... is a sequence defined by the recurrence relation f, = fr -1+ 2k for each integer k 2 2, with initial condition f, = 1.
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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![### Proof by Mathematical Induction
#### Problem Statement:
We are given a sequence defined by the recurrence relation:
\[ f_k = f_{k-1} + 2^k \text{ for each integer } k \geq 2 \]
\[ f_1 = 1 \]
We need to show that this sequence satisfies the following explicit formula:
\[ f_n = 2^{n+1} + 3 \text{ for every integer } n \geq 1 \]
#### Proof (by mathematical induction):
Suppose \( f_1, f_2, f_3, \ldots \) is a sequence that satisfies the recurrence relation
\[ f_k = f_{k-1} + 2^k \text{ for each integer } k \geq 2 \]
with initial condition \( f_1 = 1 \).
We need to show that when the sequence \( f_1, f_2, f_3, \ldots \) is defined in this recursive way, all terms in the sequence also satisfy the explicit formula shown above.
So let the property \( P(n) \) be the equation \( f_n = 2^{n+1} + 3 \). We will show that \( P(n) \) is true for every integer \( n \geq 1 \).
#### Show that \( P(1) \) is true:
The left-hand side of \( P(1) \) is \( f_1 \), which equals \( 1 \). The right-hand side of \( P(1) \) is \( 2^{1+1} + 3 \). Since the left-hand side and right-hand sides equal each other, \( P(1) \) is true.
\[
f_1 = 1 = 2^{1 + 1} + 3 = 4 + 3 = 7
\]
#### Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:
Let \( k \) be any integer with \( k \geq 1 \), and suppose that \( P(k) \) is true. In other words, suppose that \( f_k = 2^{k + 1} + 3 \), where \( f_1, f_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8268c271-8991-49a2-aed9-a02bde5bd8ab%2Fbac71982-e5c7-416a-9888-adaccf51ecdf%2Fbn6vqnq_processed.png&w=3840&q=75)
Transcribed Image Text:### Proof by Mathematical Induction
#### Problem Statement:
We are given a sequence defined by the recurrence relation:
\[ f_k = f_{k-1} + 2^k \text{ for each integer } k \geq 2 \]
\[ f_1 = 1 \]
We need to show that this sequence satisfies the following explicit formula:
\[ f_n = 2^{n+1} + 3 \text{ for every integer } n \geq 1 \]
#### Proof (by mathematical induction):
Suppose \( f_1, f_2, f_3, \ldots \) is a sequence that satisfies the recurrence relation
\[ f_k = f_{k-1} + 2^k \text{ for each integer } k \geq 2 \]
with initial condition \( f_1 = 1 \).
We need to show that when the sequence \( f_1, f_2, f_3, \ldots \) is defined in this recursive way, all terms in the sequence also satisfy the explicit formula shown above.
So let the property \( P(n) \) be the equation \( f_n = 2^{n+1} + 3 \). We will show that \( P(n) \) is true for every integer \( n \geq 1 \).
#### Show that \( P(1) \) is true:
The left-hand side of \( P(1) \) is \( f_1 \), which equals \( 1 \). The right-hand side of \( P(1) \) is \( 2^{1+1} + 3 \). Since the left-hand side and right-hand sides equal each other, \( P(1) \) is true.
\[
f_1 = 1 = 2^{1 + 1} + 3 = 4 + 3 = 7
\]
#### Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:
Let \( k \) be any integer with \( k \geq 1 \), and suppose that \( P(k) \) is true. In other words, suppose that \( f_k = 2^{k + 1} + 3 \), where \( f_1, f_
![We must show that \( P(k + 1) \) is true. In other words, we must show that \( f_{k+1} = \boxed{} \). Now the left-hand side of \( P(k + 1) \) is
\[
f_{k+1} = f_k + 2^{k+1}
\]
by \(\boxed{\text{---Select---}}\),
\[
= \boxed{} + 2^{k+1}
\]
by \(\boxed{\text{---Select---}}\),
\[
= 2 \cdot \left( \boxed{} \right) - 3
\]
\[
= \boxed{}
\]
by \(\boxed{\text{---Select---}}\),
and this is the right-hand side of \( P(k + 1) \). Hence the inductive step is complete.
*Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8268c271-8991-49a2-aed9-a02bde5bd8ab%2Fbac71982-e5c7-416a-9888-adaccf51ecdf%2Fzp1kp7_processed.png&w=3840&q=75)
Transcribed Image Text:We must show that \( P(k + 1) \) is true. In other words, we must show that \( f_{k+1} = \boxed{} \). Now the left-hand side of \( P(k + 1) \) is
\[
f_{k+1} = f_k + 2^{k+1}
\]
by \(\boxed{\text{---Select---}}\),
\[
= \boxed{} + 2^{k+1}
\]
by \(\boxed{\text{---Select---}}\),
\[
= 2 \cdot \left( \boxed{} \right) - 3
\]
\[
= \boxed{}
\]
by \(\boxed{\text{---Select---}}\),
and this is the right-hand side of \( P(k + 1) \). Hence the inductive step is complete.
*Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.*
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