rove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. n(5n - 3) For every integer n2 1, 1 + 6 + 11 + 16 +... + (5n - 4) = 2

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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 + ... + (5n - 4) = n(5n - 3)
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 n2 1.
Show that P(1) is true: Select P(1) from the choices below.
O P(1) = 5. 1 - 4
O 1 =1:(5 1- 3)
2
O 1+ (5.1- 4) = 1. (5. 1- 3)
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 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. 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--
v + (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.]
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 + ... + (5n - 4) = n(5n - 3) 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 n2 1. Show that P(1) is true: Select P(1) from the choices below. O P(1) = 5. 1 - 4 O 1 =1:(5 1- 3) 2 O 1+ (5.1- 4) = 1. (5. 1- 3) 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 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. 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-- v + (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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