Question: Use the Principle of Mathematical Induction to show that the given statement is true for all…

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \( n \).

\[ 7 + 15 + 23 + \cdots + (8n - 1) = n(4n + 3) \]

To prove that \( 7 + 15 + 23 + \cdots + (8n - 1) = n(4n + 3) \) is true for all natural numbers \( n \), first show that this statement is true for \( n = 1 \).

Simplify each side of the statement for the appropriate value of \( n \).

\[ 7 + 15 + 23 + \cdots + (8n - 1) = n(4n + 3) \]

\[ \boxed{\phantom{0}} = \boxed{\phantom{0}} \]

Solution

To prove the given statement using mathematical induction, follow these steps: Base Case: Check if the statement is true for \( n = 1 \). Substitute \( n = 1 \) into the left side of the equation: \[ 7 + 15 + 23 + \ldots + (8 \times 1 - 1) = 8 \times 1 - 1 = 7 \] Now, substitute \( n = 1 \) into the right side of the equation: \[ 1(4 \times 1 + 3) = 1 \times 7 = 7 \] Since both sides are equal, the base case is true. Inductive Step: Assume the statement is true for \( n = k \): \[ 7 + 15 + 23 + \ldots + (8k - 1) = k(4k + 3) \] Now prove it for \( n = k + 1 \): Add the next term \( 8(k + 1) - 1 \) to both sides: Left side: \[ 7 + 15 + 23 + \ldots + (8k - 1) + (8(k + 1) - 1) \] According to the inductive hypothesis, the left side becomes: \[ k(4k + 3) + (8(k + 1) - 1) \] Simplify the right side: \[ k(4k + 3) + 8k + 8 - 1 \] \[ = 4k^2 + 3k + 8k + 7 \] \[ = 4k^2 + 11k + 7 \] Right side for \( n = k + 1 \): \[ (k + 1)(4(k + 1) + 3) \] \[ = (k + 1)(4k + 4 + 3) \] \[ = (k + 1)(4k + 7) \] Expand it: \[ = k(4k + 7) + 1(4k + 7) \] \[ = 4k^2 + 7k + 4k + 7 \] \[ = 4k^2 + 11k + 7 \] Since the left side equals the right side, the inductive step is true. Therefore, by the principle of mathematical induction, the statement is true for all natural numbers \( n \).