3n Prove that for every integer n > 0, 9|(4" + 8).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Solve it by using induction method:

Transcribed Image Text:**Exercise: Mathematical Induction**
**Problem Statement:**
Prove that for every integer \( n \geq 0 \), \( 9 \mid (4^{3n} + 8) \).
**Solution Approach:**
To prove this statement, you may consider using the method of mathematical induction. The process involves the following steps:
1. **Base Case:** Verify the statement for the initial value \( n = 0 \).
2. **Inductive Step:** Assume the statement is true for \( n = k \), i.e., \( 9 \mid (4^{3k} + 8) \). Then, prove it for \( n = k + 1 \).
**Explanation:**
The notation \( 9 \mid (4^{3n} + 8) \) means that the expression \( 4^{3n} + 8 \) is divisible by 9 for every integer \( n \) greater than or equal to 0.
You will find that both steps of induction confirm the divisibility condition, solidifying the statement for all non-negative integers \( n \).
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