Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation fr = fk-1 f = 1 + 2k for each integer k 2 2 satisfies the following formula. f. = 2" + 1- 3 for every integer n 2 1 Proof (by mathematical induction): Suppose f,, f,, fy... is a sequence that satisfies the recurrence relation f, = f,+ 2k 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) is4-3 , which equals 1 . The right-hand side of P(1) is 1 Since the left-hand and right-hand sides equal each other, P(1) is true. Show that for each integer k2 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, = 21-3 where f,, f, f . is a sequence defined by the recurrence relation f = f, + 2k 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 f1 ok+2 3 . Now the left-hand side of P(k + 1) is k+1 = f + 2k +1 gk+1 by definition of f1, f2, fs, ... + 2k + 1 by inductive hypothesis - 3 k+1 = 2 = ok+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.)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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What is the answer for the first one? And please explain why is it wrong? My teacher said “They are not looking for a number” and I don’t really understand that. Does that mean I need to put the f1 = in front of the answer or something?
Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation
fk = fk -1
f = 1
+ 2k for each integer k 2 2
satisfies the following formula.
f. = 2" +1- 3 for every integer n 21
Proof (by mathematical induction):
Suppose f,, f,, fy... is a sequence that satisfies the recurrence relation
f, = f, + 2k for each integer k 2 2, with initial condition f, = 1.
We need to show that when the sequence f,, f, fay... 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 4-3
, which equals 1
. The right-hand side of P(1) is 1
Since the left-hand and right-hand sides equal each other, P(1) is true.
Show that for each integer k2 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, = 21-3
where f,, f,fa.. is a sequence defined by the recurrence relation
f = f,+ 2k 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 f1
k+2
3
. Now the left-hand side of P(k + 1) is
%3D
by definition of f1, f2, fs, ...
k+1 = f + 2k + 1
ok+1
3
+ 2k + 1
by inductive hypothesis
k+1
= 2
= ok+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.)
Transcribed Image Text:Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation fk = fk -1 f = 1 + 2k for each integer k 2 2 satisfies the following formula. f. = 2" +1- 3 for every integer n 21 Proof (by mathematical induction): Suppose f,, f,, fy... is a sequence that satisfies the recurrence relation f, = f, + 2k for each integer k 2 2, with initial condition f, = 1. We need to show that when the sequence f,, f, fay... 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 4-3 , which equals 1 . The right-hand side of P(1) is 1 Since the left-hand and right-hand sides equal each other, P(1) is true. Show that for each integer k2 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, = 21-3 where f,, f,fa.. is a sequence defined by the recurrence relation f = f,+ 2k 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 f1 k+2 3 . Now the left-hand side of P(k + 1) is %3D by definition of f1, f2, fs, ... k+1 = f + 2k + 1 ok+1 3 + 2k + 1 by inductive hypothesis k+1 = 2 = ok+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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