y mathematical induction): f, f2, f3,... is a sequence that satisfies the recurrence relation f, = f, - + 2 for each integer k 2 2, with initial condition f, = 1. d 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. he property P(n) be the equation f = 2" + - - 3. We will show that P(n) is true for every integer n 2 1. hat P(1) is true: t-hand side of P(1) is 2° -3 which equals 1 The right-hand side of P(1) is 1 Since the left-hand and right- hat for each integer k 2 1, if P(k) is true, then P(k + 1) is true: e any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = |クk+1-3 where f,, f,, fa,... is a sequence defined by the r fy = f -1+ 2* for each integer k 2 2, with initial condition f, = 1. s the inductive hypothesis.] ust show that P(k + 1) is true. In other words, we must show that f, *+2 -3 + 1 Now the left-hand side of P(k + 1) is k +1 = f, + 2k +1 by definition of f., fa, f, .v 2k +1 - 3 + 2k + 1 %3D by the laws of algebra 2k+1-3 = 2. - 3 2k+2-3 by the laws of algebra this is the right-hand side of P(k + 1). Hence the inductive step is complete. "s, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete 1
y mathematical induction): f, f2, f3,... is a sequence that satisfies the recurrence relation f, = f, - + 2 for each integer k 2 2, with initial condition f, = 1. d 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. he property P(n) be the equation f = 2" + - - 3. We will show that P(n) is true for every integer n 2 1. hat P(1) is true: t-hand side of P(1) is 2° -3 which equals 1 The right-hand side of P(1) is 1 Since the left-hand and right- hat for each integer k 2 1, if P(k) is true, then P(k + 1) is true: e any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = |クk+1-3 where f,, f,, fa,... is a sequence defined by the r fy = f -1+ 2* for each integer k 2 2, with initial condition f, = 1. s the inductive hypothesis.] ust show that P(k + 1) is true. In other words, we must show that f, *+2 -3 + 1 Now the left-hand side of P(k + 1) is k +1 = f, + 2k +1 by definition of f., fa, f, .v 2k +1 - 3 + 2k + 1 %3D by the laws of algebra 2k+1-3 = 2. - 3 2k+2-3 by the laws of algebra this is the right-hand side of P(k + 1). Hence the inductive step is complete. "s, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete 1
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
For the left hand side of p(1) they are not looking for number but they are looking for something in term of f so what is the answer? Should it be f1?
![f = 2 + 1
3 for every integer n 2 1
(by mathematical induction):
se f,, f2, f3 is a sequence that satisfies the recurrence relation
f, = f, -1 + 2 for each integer k 2 2, with initial condition f, = 1.
eed to show that when the sequence f,, f,, f3
is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above.
t the property P(n) be the equation f = 2" + - 3. We will show that P(n) is true for every integer n 2 1.
- that P(1) is true:
eft-hand side of P(1) is 2
- 3
, which equals 1
The right-hand side of P(1) is 1
Since the left-hand and right-
v that for each integer k 2 1, if P(k) is true, then P(k + 1) is true:
R be any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = 9+1
3
where f,, f2, f3
... is a sequence defined by the re
f, = f, -1 + 2* for each integer k 2 2, with initial condition f, = 1.
s is the inductive hypothesis.]
must show that P(k + 1) is true. In other words, we must show that f,
k +2
Now the left-hand side of P(k + 1) is
k +1 = f, + 2k +1
by definition of f, fa, f.
gk +1
- 3
+ 2k + 1
by the laws of algebra
= 2 -
っk+1
- 3
- 3
gk+2-3
%3D
by the laws of algebra
d 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 comniete 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9ee0bdb-6e61-488d-8ad0-e9784396cf1d%2F93edfb94-2f7f-4fed-862b-4184bb77093a%2Fc927gl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:f = 2 + 1
3 for every integer n 2 1
(by mathematical induction):
se f,, f2, f3 is a sequence that satisfies the recurrence relation
f, = f, -1 + 2 for each integer k 2 2, with initial condition f, = 1.
eed to show that when the sequence f,, f,, f3
is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above.
t the property P(n) be the equation f = 2" + - 3. We will show that P(n) is true for every integer n 2 1.
- that P(1) is true:
eft-hand side of P(1) is 2
- 3
, which equals 1
The right-hand side of P(1) is 1
Since the left-hand and right-
v that for each integer k 2 1, if P(k) is true, then P(k + 1) is true:
R be any integer with k 2 1, and suppose that P(k) is true. In other words, suppose that f, = 9+1
3
where f,, f2, f3
... is a sequence defined by the re
f, = f, -1 + 2* for each integer k 2 2, with initial condition f, = 1.
s is the inductive hypothesis.]
must show that P(k + 1) is true. In other words, we must show that f,
k +2
Now the left-hand side of P(k + 1) is
k +1 = f, + 2k +1
by definition of f, fa, f.
gk +1
- 3
+ 2k + 1
by the laws of algebra
= 2 -
っk+1
- 3
- 3
gk+2-3
%3D
by the laws of algebra
d 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 comniete 1
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