7n - 2" is divisible by 5. We will show that P(n) is true for every integer n 2 0. Show that P(0) is true: Select P(0) from the choices below. O (70 – 20) | 5 O 5 is a multiple of 70 - 20 O 70 - 20 < 5 OS (70 - 20) The truth of the selected statement follows from the definition of divisibility and the fact that 70- ? v- 0. Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. Select P(k) from the choices below. O 5 is a multiple of 7k - 2k O zk - 2k < 5 O zk - 2k is divisible by 5 O 5 is divisible by 7k - 2k [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 5 is divisible by 7k +1- 2k +1 O zk+1-2k+1<5 O s is a multiple of 7k + 1- 2k +1 O zk +1- 2k+ 1 is divisible by 5 By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 2* - 5r. Then 7k +1- 2k+1-7. 7k - 2-2k- (5 + 2) - 7k - 2.2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k +1- 2k+15.
7n - 2" is divisible by 5. We will show that P(n) is true for every integer n 2 0. Show that P(0) is true: Select P(0) from the choices below. O (70 – 20) | 5 O 5 is a multiple of 70 - 20 O 70 - 20 < 5 OS (70 - 20) The truth of the selected statement follows from the definition of divisibility and the fact that 70- ? v- 0. Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. Select P(k) from the choices below. O 5 is a multiple of 7k - 2k O zk - 2k < 5 O zk - 2k is divisible by 5 O 5 is divisible by 7k - 2k [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 5 is divisible by 7k +1- 2k +1 O zk+1-2k+1<5 O s is a multiple of 7k + 1- 2k +1 O zk +1- 2k+ 1 is divisible by 5 By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 2* - 5r. Then 7k +1- 2k+1-7. 7k - 2-2k- (5 + 2) - 7k - 2.2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k +1- 2k+15.
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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![Prove the following statenment by mathematical induction.
For every integer n 2 0, 7" - 2" is divisible by 5.
Proof (by mathematical induction): Let P(n) be the following sentence.
7n - 2" is divisible by 5.
We will show that P(n) is true for every integer n 2 0.
Show that P(0) is true: Select P(0) from the choices below,
O (70 – 20) I 5
O 5 is a multiple of 70 - 20
O 70 - 2° < 5
O 5 |(7° – 2º)
The truth of the selected statement follows from the definition of divisibility and the fact that 70-? v- 0.
Show that for each integer k > 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. Select P(k) from the choices below.
O 5 is a multiple of 7k - 2k
O zk - 2k < 5
O zk - 2k is divisible by 5
O 5 is divisible by 7k - 2k
[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 is divisible by 7k + 1 - 2k+1
O zk +1 - 2k +1<5
O 5 is a multiple of 7k + 1 _ 2k +1
O zk +1 _ 2k+ 1 is divisible by 5
By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k- 2k - 5r. Then
zk +1 _ 2k + 1 - 7.7k - 2 - 2k = (5 + 2) · 7k – 2 · 2k.
Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r.
7k + 1_ 2k+1 5.
This quantity is an integer because k and r are integers.
Thus 7k + 1- 2k +1Select--v divisible by 5, and hence P(k + 1) is -Selectv, which completes the inductive step.
[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%2F2863cec8-c019-46a6-ad68-be87d1cedb5d%2F0fee7297-1583-4f92-b3bf-a9299afbed6a%2Fvtcskei_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Prove the following statenment by mathematical induction.
For every integer n 2 0, 7" - 2" is divisible by 5.
Proof (by mathematical induction): Let P(n) be the following sentence.
7n - 2" is divisible by 5.
We will show that P(n) is true for every integer n 2 0.
Show that P(0) is true: Select P(0) from the choices below,
O (70 – 20) I 5
O 5 is a multiple of 70 - 20
O 70 - 2° < 5
O 5 |(7° – 2º)
The truth of the selected statement follows from the definition of divisibility and the fact that 70-? v- 0.
Show that for each integer k > 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. Select P(k) from the choices below.
O 5 is a multiple of 7k - 2k
O zk - 2k < 5
O zk - 2k is divisible by 5
O 5 is divisible by 7k - 2k
[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 is divisible by 7k + 1 - 2k+1
O zk +1 - 2k +1<5
O 5 is a multiple of 7k + 1 _ 2k +1
O zk +1 _ 2k+ 1 is divisible by 5
By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k- 2k - 5r. Then
zk +1 _ 2k + 1 - 7.7k - 2 - 2k = (5 + 2) · 7k – 2 · 2k.
Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r.
7k + 1_ 2k+1 5.
This quantity is an integer because k and r are integers.
Thus 7k + 1- 2k +1Select--v divisible by 5, and hence P(k + 1) is -Selectv, which completes the inductive step.
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
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