
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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)
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> 1.
Show that P(1) is true: Select P(1) from the choices below.
O1=1: (5 - 1 - 3)
2
O P(1) = 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 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-
, and the right-hand side of P(k) is
[The inductive hypothesis states that the two sides of P(k) are equal.]
After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes
We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + - + (5(k + 1) - 4) =
Hence P(k + 1) is true, which completes the inductive step.
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
[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)
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> 1.
Show that P(1) is true: Select P(1) from the choices below.
O1=1: (5 - 1 - 3)
2
O P(1) = 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 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-
, and the right-hand side of P(k) is
[The inductive hypothesis states that the two sides of P(k) are equal.]
After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes
We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + - + (5(k + 1) - 4) =
Hence P(k + 1) is true, which completes the inductive step.
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
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
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