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The problem is asking us to use mathematical induction to prove that the expression \(10^n + 3 \cdot 4^{n+2} + 5\) is divisible by 9 for all natural numbers \(n\).

### Step 1: Base Case
For \(n = 1\):

\[
10^1 + 3 \cdot 4^{1+2} + 5 = 10 + 3 \cdot 4^3 + 5 = 10 + 3 \cdot 64 + 5 = 10 + 192 + 5 = 207
\]

Now, check if 207 is divisible by 9:

\[
207 \div 9 = 23
\]

Since 207 is divisible by 9, the base case holds.

### Step 2: Inductive Step
Assume the statement is true for \(n = k\); that is, assume:

\[
10^k + 3 \cdot 4^{k+2} + 5 \text{ is divisible by 9}
\]

This means:

\[
10^k + 3 \cdot 4^{k+2} + 5 = 9m \quad \text{for some integer } m.
\]

### Step 3: Prove for \(n = k + 1\)
We need to prove that:

\[
10^{k+1} + 3 \cdot 4^{(k+1) + 2} + 5 \text{ is divisible by 9}.
\]

Simplify the expression:

\[
10^{k+1} + 3 \cdot 4^{k+3} + 5
\]

Rewrite \(10^{k+1}\) as \(10 \cdot 10^k\):

\[
10 \cdot 10^k + 3 \cdot 4^{k+3} + 5
\]

Substituting the inductive hypothesis (\(10^k + 3 \cdot 4^{k+2} + 5 = 9m\)):

\[
= 10 \cdot (9m - 3 \cdot 4^{k+2} - 5) + 3 \cdot 4^{k+3} + 5
\]

Simplify further:

\[
= 90m - 30 \cdot 4^{k+2} - 50 + 3 \cdot 4^{k+3} + 5
\]

Notice \(4^{k+3} = 4 \cdot 4^{k+2}\):

\[
= 90m - 30 \cdot 4^{k+2} - 50 + 12 \cdot 4^{k+2} + 5
\]

Simplify the expression:

\[
= 90m - 18 \cdot 4^{k+2} - 45
\]

Finally, we get:

\[
= 9 (10m - 2 \cdot 4^{k+2} - 5)
\]

Since the expression is in the form of \(9 \times \text{integer}\), it is divisible by 9. This completes the inductive step.

Therefore, by the principle of mathematical induction, \(10^n + 3 \cdot 4^{n+2} + 5\) is divisible by 9 for all natural numbers \(n\).

Would you like more details, or do you have any specific questions? 

Here are 5 related questions to explore:
1. How does mathematical induction work in proving divisibility problems?
2. Can we use other methods to prove that the expression is divisible by 9?
3. What are some common mistakes when applying mathematical induction?
4. How does changing the base case affect the induction process?
5. Can mathematical induction be used to prove inequalities?

**Tip:** When using mathematical induction, ensure your base case is correct and your inductive step logically follows the assumption.

Use the mathematical induction principle to prove that 10^n + 3 * 4^(n+2) + 5 is divisible by 9 for all natural numbers n.

Learn how to use the principle of mathematical induction to prove that 10^n + 3 * 4^(n+2) + 5 is divisible by 9 for all natural numbers n. This step-by-step solution covers key concepts like exponents and divisibility rules. Ideal for high school students.

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