For every integer n2 1, 1+6+ 11 + 16 + + (Sn - 4) = Sn - 3) Proof (by mathematical induction): Let P(n) be the equation 1+6+ 11 + 16 + + (Sn - 4) - (Sn-3) 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. O1-1- (5-1-3) O P(1) = 1- (5-1- 3) O P(1) - 5.1-4 O1+ (5.1-4) = 1. (5-1-3) The selected statement is true because both sides of the equation equal Show that for each integer k21, 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-- and the right-hand side of P(k) is [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) = After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes Select- + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 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.]
For every integer n2 1, 1+6+ 11 + 16 + + (Sn - 4) = Sn - 3) Proof (by mathematical induction): Let P(n) be the equation 1+6+ 11 + 16 + + (Sn - 4) - (Sn-3) 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. O1-1- (5-1-3) O P(1) = 1- (5-1- 3) O P(1) - 5.1-4 O1+ (5.1-4) = 1. (5-1-3) The selected statement is true because both sides of the equation equal Show that for each integer k21, 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-- and the right-hand side of P(k) is [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) = After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes Select- + (5(k + 1) - 4). When the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 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.]
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.
For every integer n2 1, 1 +6 + 11 + 16 + ... + (Sn - 4) = (Sn - 3)
Proof (by mathematical induction): Let P(n) be the equation
1+6+ 11 + 16 + + (Sn - 4) = (Sn- 3)
We will show that P(n)-is true for every integer n 21.
Show that P(1) is true: Select P(1) from the choices below.
O1= 1: (5. 1- 3)
O P) a 1. (5.1- 3)
2
O P(1) = 5.1- 4
O1+ (5-1 - 4) = 1. (5-1- 3)
The selected statement is true because both sides of the equation equal
Show that for each integer kz 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-
V, and the right-hand side of P(k) is
[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) =
After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes Select--
+ (5(k + 1) - 4). When the
left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal
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.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce233c24-0412-41c5-a669-703bdad8237e%2F51b46d20-9741-4f3d-ba12-1404bde7ec2e%2Fm9xlhs_processed.jpeg&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.
For every integer n2 1, 1 +6 + 11 + 16 + ... + (Sn - 4) = (Sn - 3)
Proof (by mathematical induction): Let P(n) be the equation
1+6+ 11 + 16 + + (Sn - 4) = (Sn- 3)
We will show that P(n)-is true for every integer n 21.
Show that P(1) is true: Select P(1) from the choices below.
O1= 1: (5. 1- 3)
O P) a 1. (5.1- 3)
2
O P(1) = 5.1- 4
O1+ (5-1 - 4) = 1. (5-1- 3)
The selected statement is true because both sides of the equation equal
Show that for each integer kz 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-
V, and the right-hand side of P(k) is
[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) =
After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes Select--
+ (5(k + 1) - 4). When the
left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal
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.]
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