Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer n 2 1,1 + 6 + 11 + 16 + + (5n - 4) = n(5n-3) 2 Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + + (5n-4)= n(Sn - 3) 2 We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: Select P(1) from the choices below. O1+ (5 1-4)= 1·(5·1-3) O P(1) 5 1-4 0 1 = 1 (5-1-3) 2 O P(1) = 1·(5·1-3) 2 The selected statement is true because both sides of the equation equal 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. The left-hand side of P(k) is ---Select--- [The inductive hypothesis states that the two sides of P(k) are equal.] We must show that P(k+ 1) is true. P(k+ 1) is the equation 1 + 6 + 11 + 16 ++ (5(k+ 1) − 4) = the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal ✓, and the right-hand side of P(k) is After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes ---Select--- . Hence P(k+ 1) is true, 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.] +(5(k+1)-4). When
Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer n 2 1,1 + 6 + 11 + 16 + + (5n - 4) = n(5n-3) 2 Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + + (5n-4)= n(Sn - 3) 2 We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: Select P(1) from the choices below. O1+ (5 1-4)= 1·(5·1-3) O P(1) 5 1-4 0 1 = 1 (5-1-3) 2 O P(1) = 1·(5·1-3) 2 The selected statement is true because both sides of the equation equal 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. The left-hand side of P(k) is ---Select--- [The inductive hypothesis states that the two sides of P(k) are equal.] We must show that P(k+ 1) is true. P(k+ 1) is the equation 1 + 6 + 11 + 16 ++ (5(k+ 1) − 4) = the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal ✓, and the right-hand side of P(k) is After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes ---Select--- . Hence P(k+ 1) is true, 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.] +(5(k+1)-4). When
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 statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2.
n(5n - 3)
2
For every integer n ≥ 1, 1 + 6 + 11 + 16 + ... + (5n - 4) =
Proof (by mathematical induction): Let P(n) be the equation
n(5n - 3)
2
1 + 6 + 11 + 16 + + (5n - 4) =
...
We will show that P(n) is true for every integer n 2 1.
Show that P(1) is true: Select P(1) from the choices below.
O 1 + (514) = 1·(5·1-3)
O P(1) = 5.14
0 1 = 1. (5.1-3)
2
O P(1) =
1. (5.1 - 3)
2
The selected statement is true because both sides of the equation equal
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 ≥ 1, and suppose that P(k) is true. The left-hand side of P(k) is ---Select---
[The inductive hypothesis states that the two sides of P(k) are equal.]
We must show that P(k+ 1) is true. P(k+ 1) is the equation 1 + 6 + 11 + 16 + ... + (5(k + 1) − 4) =
the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal
and the right-hand side of P(k) is
After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes --Select---
Hence P(x + 1) is true, 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.]
✓+ (5(k+ 1) - 4). When](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7cff8c4a-bb33-4b54-bd2d-8ccd43ef3f82%2F3bf17f46-e89a-4fc9-a75e-2874b52e6dad%2Fkh538f_processed.png&w=3840&q=75)
Transcribed Image Text:Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2.
n(5n - 3)
2
For every integer n ≥ 1, 1 + 6 + 11 + 16 + ... + (5n - 4) =
Proof (by mathematical induction): Let P(n) be the equation
n(5n - 3)
2
1 + 6 + 11 + 16 + + (5n - 4) =
...
We will show that P(n) is true for every integer n 2 1.
Show that P(1) is true: Select P(1) from the choices below.
O 1 + (514) = 1·(5·1-3)
O P(1) = 5.14
0 1 = 1. (5.1-3)
2
O P(1) =
1. (5.1 - 3)
2
The selected statement is true because both sides of the equation equal
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 ≥ 1, and suppose that P(k) is true. The left-hand side of P(k) is ---Select---
[The inductive hypothesis states that the two sides of P(k) are equal.]
We must show that P(k+ 1) is true. P(k+ 1) is the equation 1 + 6 + 11 + 16 + ... + (5(k + 1) − 4) =
the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal
and the right-hand side of P(k) is
After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes --Select---
Hence P(x + 1) is true, 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.]
✓+ (5(k+ 1) - 4). When
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