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We are tasked with proving that the expression \( 6^{k+2} + 6^{k+1} - 18(6^k) \) is divisible by 24 for all positive integer values of \( k \).

Let's analyze and simplify the given expression step by step.

### Step 1: Factor the expression
We observe that each term in the expression contains a factor of \( 6^k \), so we can factor that out:
\[
6^{k+2} + 6^{k+1} - 18(6^k) = 6^k \left(6^2 + 6^1 - 18\right)
\]

### Step 2: Simplify the terms inside the parentheses
Now, simplify the terms inside the parentheses:
\[
6^2 = 36, \quad 6^1 = 6
\]
Substituting these values gives:
\[
6^k \left(36 + 6 - 18\right) = 6^k \left(24\right)
\]

Thus, the expression simplifies to:
\[
6^k \cdot 24
\]

### Step 3: Conclusion
Since the expression is \( 6^k \cdot 24 \), it is clear that it is divisible by 24 for any positive integer \( k \), because 24 is a factor of the entire expression.

Thus, we have shown that \( 6^{k+2} + 6^{k+1} - 18(6^k) \) is divisible by 24 for all positive integer values of \( k \).

Let me know if you'd like any further clarification!

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### 5 Related Questions:
1. How would you generalize this result for different bases instead of 6?
2. What are other methods for proving divisibility by a specific number?
3. Can you apply this factorization approach to expressions with other powers, like \( 7^k \)?
4. How does this problem relate to modular arithmetic?
5. Can you derive other divisibility properties using this method?

### Tip:
When proving divisibility, always check if factoring simplifies the expression, as this often reveals common divisors easily.

Show that 6^{k+2} + 6^{k+1} - 18(6^k) is divisible by 24 for all positive integer values of k.

Learn how to prove that the expression 6^{k+2} + 6^{k+1} - 18(6^k) is divisible by 24 for all positive integer values of k. This solution covers key concepts like algebraic factorization and divisibility techniques, suitable for students in Grades 10-12.

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