Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation k = -1 + 2* for each integer k 2 2 f, = 1 satisfies the following formula. = 2n +1- 3 for every integer n 2 1 Proof (by mathematical induction): Suppose f,, f, fa,... is a sequence that satisfies the recurrence relation f, = f,-1 + 2* for each integer k 2 2, with initial condition f, = 1. We need to show that when the sequence f,, f, fa, ... 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 integerk2 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,, f2, f3 ... is a sequence defined by the recurrence relation f, = f,- + 2* for each integer k 2 2, with initial condition f, = 1. [This is the inductive hypothesis.) We must show that P(k + 1) is true. In other words, we must show that f, 41 = Now the left-hand side of P(k + 1) is k +1 = f, + 2* + 1 by definition of f, fe, fi, + 2k +1 by inductive hypothesis v %3D = 2 - by the laws of algebra 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.]
Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation k = -1 + 2* for each integer k 2 2 f, = 1 satisfies the following formula. = 2n +1- 3 for every integer n 2 1 Proof (by mathematical induction): Suppose f,, f, fa,... is a sequence that satisfies the recurrence relation f, = f,-1 + 2* for each integer k 2 2, with initial condition f, = 1. We need to show that when the sequence f,, f, fa, ... 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 integerk2 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,, f2, f3 ... is a sequence defined by the recurrence relation f, = f,- + 2* for each integer k 2 2, with initial condition f, = 1. [This is the inductive hypothesis.) We must show that P(k + 1) is true. In other words, we must show that f, 41 = Now the left-hand side of P(k + 1) is k +1 = f, + 2* + 1 by definition of f, fe, fi, + 2k +1 by inductive hypothesis v %3D = 2 - by the laws of algebra 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.]
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.2: Mathematical Induction
Problem 53E: Given the recursively defined sequence a1=0,a2=30, and an=8an115an2, use complete induction to prove...
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![Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation
fk = f,
f, = 1
+ 2* for each integer k 2 2
- 1
satisfies the following formula.
f. = 2" +1 - 3 for every integer n 2 1
Proof (by mathematical induction):
Suppose f,, fa, fa,... is a sequence that satisfies the recurrence relation
f, = f, -1+ 2* for each integer k 2 2, with initial condition f, = 1
We need to show that when the sequence f,, f, f, ... 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" +- - 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 integerk2 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, fa,... is a sequence defined by the recurrence relation
f, = f, -, + 2* for each integer k 2 2, with initial condition f, = 1
[This is the inductive hypothesis.]
We must show that P(k + 1) is true. In other words, we must show that f, 1 =
Now the left-hand side of P(k + 1) is
by definition of fi, fa, fs, .
'k +1 = f, + 2k + 1
+ 2k +1
by inductive hypothesis v
= 2.
·3
by the laws of algebra
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%2Fd197b407-d385-4d92-989e-57a34001c894%2Fffd70939-0f99-469d-95ef-7d4cdfc2d53d%2Fcckuqr9_processed.png&w=3840&q=75)
Transcribed Image Text:Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation
fk = f,
f, = 1
+ 2* for each integer k 2 2
- 1
satisfies the following formula.
f. = 2" +1 - 3 for every integer n 2 1
Proof (by mathematical induction):
Suppose f,, fa, fa,... is a sequence that satisfies the recurrence relation
f, = f, -1+ 2* for each integer k 2 2, with initial condition f, = 1
We need to show that when the sequence f,, f, f, ... 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" +- - 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 integerk2 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, fa,... is a sequence defined by the recurrence relation
f, = f, -, + 2* for each integer k 2 2, with initial condition f, = 1
[This is the inductive hypothesis.]
We must show that P(k + 1) is true. In other words, we must show that f, 1 =
Now the left-hand side of P(k + 1) is
by definition of fi, fa, fs, .
'k +1 = f, + 2k + 1
+ 2k +1
by inductive hypothesis v
= 2.
·3
by the laws of algebra
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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