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++ (Sn-4)= n(n-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 n 2 1. Show that P(1) is true: Select P(1) from the choices below. 01+(5-1-4) 1 (5-1- 3) 1-1-(5-1-3) 2 O P(1) = 1·(5 - 1-(5-1-3) 2 OP(1)-5-1-4 The selected statement is true because both sides of the equation equal 1 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 2 1, and suppose that P(K) is true. The left-hand side of P(K) is 1 + 6 + 11 + 16 + (Sk-4) X, and the right-hand side of P(k) is 1 [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)= +6+11+16++ right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 1 (k+ 1) (5(k+1)-3) 2 Hence P(k+ 1) is true, which completes the inductive step. X [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] x After substitution from the inductive hypothesis, the left-hand side of P(x + 1) becomes (k(5k - 3)/2 ✔+(5(k+1)-4). When the left-hand and
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++ (Sn-4)= n(n-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 n 2 1. Show that P(1) is true: Select P(1) from the choices below. 01+(5-1-4) 1 (5-1- 3) 1-1-(5-1-3) 2 O P(1) = 1·(5 - 1-(5-1-3) 2 OP(1)-5-1-4 The selected statement is true because both sides of the equation equal 1 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 2 1, and suppose that P(K) is true. The left-hand side of P(K) is 1 + 6 + 11 + 16 + (Sk-4) X, and the right-hand side of P(k) is 1 [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)= +6+11+16++ right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 1 (k+ 1) (5(k+1)-3) 2 Hence P(k+ 1) is true, which completes the inductive step. X [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] x After substitution from the inductive hypothesis, the left-hand side of P(x + 1) becomes (k(5k - 3)/2 ✔+(5(k+1)-4). When the left-hand and
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
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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 ≥ 1, 1 + 6 + 11 + 16 +... + (5n
4)
n(5n - 3)
2
Proof (by mathematical induction): Let P(n) be the equation
n(5n - 3)
2
We will show that P(n) is true for every integer n ≥ 1.
Show that P(1) is true: Select P(1) from the choices below.
O 1+(5 14) = 1. (5.1 -
3)
01=
1
1 + 6 + 11 + 16 + + (5n - 4)
=
O P(1) =
(513)
2
1. (5.13)
2
O P(1) = 5.1-4
The selected statement is true because both sides of the equation equal 1
Show that for each integer k ≥ 1, if P(k) is true, then P(k+ 1) is true:
1
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) ✓X , 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) =
right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 1
(k+ 1) (5(k+1) − 3)
2
. After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes k(5k - 3)/2
. Hence P(k + 1) is true, which completes the inductive step.
X
[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 the left-hand and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3cf874b-7a7b-478f-a7b8-421442e72224%2Fdb66b7e9-dade-4184-a62e-923aff71f6f3%2F2ngyq2_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.
For every integer n ≥ 1, 1 + 6 + 11 + 16 +... + (5n
4)
n(5n - 3)
2
Proof (by mathematical induction): Let P(n) be the equation
n(5n - 3)
2
We will show that P(n) is true for every integer n ≥ 1.
Show that P(1) is true: Select P(1) from the choices below.
O 1+(5 14) = 1. (5.1 -
3)
01=
1
1 + 6 + 11 + 16 + + (5n - 4)
=
O P(1) =
(513)
2
1. (5.13)
2
O P(1) = 5.1-4
The selected statement is true because both sides of the equation equal 1
Show that for each integer k ≥ 1, if P(k) is true, then P(k+ 1) is true:
1
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) ✓X , 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) =
right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 1
(k+ 1) (5(k+1) − 3)
2
. After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes k(5k - 3)/2
. Hence P(k + 1) is true, which completes the inductive step.
X
[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 the left-hand and
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