Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer aa 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) - A(Sn - 3) 2 We will show that Pn) is true for every integer nz 1. Show that P(1) is true: Select P1) from the choices below. 1-1 (5 1- 3) 2 01+ (5-1-4) -1. (5.1- 3) O M1) -S1-4 1. (5.1-3) O M1) -. 2 The selected statement is true because both sides of the equation equal Show that for each integer k2 1, P(K) is true, then P(k+ 1) is true: Let k be any integer with k2 1, and suppose that Pk) is true. The left-hand side of PK) is D R-4V and the right-hand side of PA) is [The inductive hypochesis states thae the two sides of PK) are equal.) We must show that Pk + 1) is true. MA 1) is the equation +6+ 11 + 16 + ** + (5(k + 1) - 4) - After substitution from the inductive hypothesis, the left-hand side of Pk+ 1) becomes S V + (5(k+ 1) - 4). When the left- true, which completes the inductive step. [Thus boch the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.) Need Help? Red
Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer aa 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) - A(Sn - 3) 2 We will show that Pn) is true for every integer nz 1. Show that P(1) is true: Select P1) from the choices below. 1-1 (5 1- 3) 2 01+ (5-1-4) -1. (5.1- 3) O M1) -S1-4 1. (5.1-3) O M1) -. 2 The selected statement is true because both sides of the equation equal Show that for each integer k2 1, P(K) is true, then P(k+ 1) is true: Let k be any integer with k2 1, and suppose that Pk) is true. The left-hand side of PK) is D R-4V and the right-hand side of PA) is [The inductive hypochesis states thae the two sides of PK) are equal.) We must show that Pk + 1) is true. MA 1) is the equation +6+ 11 + 16 + ** + (5(k + 1) - 4) - After substitution from the inductive hypothesis, the left-hand side of Pk+ 1) becomes S V + (5(k+ 1) - 4). When the left- true, which completes the inductive step. [Thus boch the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.) Need Help? Red
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 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(5n - 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.
O 1 =
1. (5· 1 - 3)
2
O 1 + (5 · 1 - 4) = 1· (5· 1 - 3)
O P(1) = 5· 1 - 4
1. (5· 1- 3)
O P(1) = -
The selected statement is true because both sides of the equation equal
Show that for each integer k 2 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 1+6 + 11 + 16 + ... + (5k
4) 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 k(5k - 3)/2
+ (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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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 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(5n - 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.
O 1 =
1. (5· 1 - 3)
2
O 1 + (5 · 1 - 4) = 1· (5· 1 - 3)
O P(1) = 5· 1 - 4
1. (5· 1- 3)
O P(1) = -
The selected statement is true because both sides of the equation equal
Show that for each integer k 2 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 1+6 + 11 + 16 + ... + (5k
4) 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 k(5k - 3)/2
+ (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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