Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Statements and Logic
Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.
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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 \( n \geq 1, 1 + 6 + 11 + 16 + \ldots + (5n - 4) = \frac{n(5n - 3)}{2} \). **Proof (by mathematical induction):** Let \( P(n) \) be the equation \[ 1 + 6 + 11 + 16 + \ldots + (5n - 4) = \frac{n(5n - 3)}{2} \] We will show that \( P(n) \) is true for every integer \( n \geq 1 \). **Show that \( P(1) \) is true:** Select \( P(1) \) from the choices below. - \( P(1) = 5 \cdot 1 - 4 \) - \( 1 = \frac{1 \cdot (5 \cdot 1 - 3)}{2} \) - \( 1 + (5 \cdot 1 - 4) = 1 \cdot (5 \cdot 1 - 3) \) - \( P(1) = 1 \cdot \frac{(5 \cdot 1 - 3)}{2} \) The selected statement is true because both sides of the equation equal \_\_\_\_\_\_\_\_\_\_. **Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:** Let \( k \) be any integer with \( k \geq 1 \), and suppose that \( P(k) \) is true. The left-hand side of \( P(k) \) is \_\_\_\_\_\_\_\_\_\_, 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
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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 \( n \geq 1, 1 + 6 + 11 + 16 + \ldots + (5n - 4) = \frac{n(5n - 3)}{2} \). **Proof (by mathematical induction):** Let \( P(n) \) be the equation \[ 1 + 6 + 11 + 16 + \ldots + (5n - 4) = \frac{n(5n - 3)}{2} \] We will show that \( P(n) \) is true for every integer \( n \geq 1 \). **Show that \( P(1) \) is true:** Select \( P(1) \) from the choices below. - \( P(1) = 5 \cdot 1 - 4 \) - \( 1 = \frac{1 \cdot (5 \cdot 1 - 3)}{2} \) - \( 1 + (5 \cdot 1 - 4) = 1 \cdot (5 \cdot 1 - 3) \) - \( P(1) = 1 \cdot \frac{(5 \cdot 1 - 3)}{2} \) The selected statement is true because both sides of the equation equal \_\_\_\_\_\_\_\_\_\_. **Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:** Let \( k \) be any integer with \( k \geq 1 \), and suppose that \( P(k) \) is true. The left-hand side of \( P(k) \) is \_\_\_\_\_\_\_\_\_\_, 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
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