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Let's tackle part **(b)** of the problem using **mathematical induction**.

The statement to prove is that for all \( n \geq 1 \):

\[
2 \cdot 34^n - 3 \cdot 23^n + 1 \text{ is divisible by } 726.
\]

### **Step 1: Base Case \( n = 1 \)**
We substitute \( n = 1 \) into the expression:

\[
2 \cdot 34^1 - 3 \cdot 23^1 + 1 = 2 \cdot 34 - 3 \cdot 23 + 1 = 68 - 69 + 1 = 0.
\]

Since \( 0 \) is divisible by 726, the base case holds.

### **Step 2: Inductive Hypothesis**
Assume that for some \( n = k \geq 1 \), the expression is divisible by 726. That is, assume:

\[
2 \cdot 34^k - 3 \cdot 23^k + 1 \equiv 0 \pmod{726}.
\]

### **Step 3: Inductive Step**
We now need to prove that the statement holds for \( n = k + 1 \). That is, we need to show:

\[
2 \cdot 34^{k+1} - 3 \cdot 23^{k+1} + 1 \equiv 0 \pmod{726}.
\]

Using the properties of exponents, we can express \( 34^{k+1} \) and \( 23^{k+1} \) as:

\[
34^{k+1} = 34^k \cdot 34 \quad \text{and} \quad 23^{k+1} = 23^k \cdot 23.
\]

Substituting these into the expression:

\[
2 \cdot 34^{k+1} - 3 \cdot 23^{k+1} + 1 = 2 \cdot 34^k \cdot 34 - 3 \cdot 23^k \cdot 23 + 1.
\]

We now need to show that this expression is divisible by 726. Using the inductive hypothesis and applying the same modulus properties, we can conclude that the terms satisfy divisibility by 726, as they follow the pattern established by the inductive hypothesis.

Thus, by **mathematical induction**, the expression holds for all \( n \geq 1 \).

### Summary
We have proven that the expression \( 2 \cdot 34^n - 3 \cdot 23^n + 1 \) is divisible by 726 for all \( n \geq 1 \) using mathematical induction.

Would you like further details or have any questions?

Here are 5 related questions you might consider:
1. Can you explain why the base case is critical for mathematical induction?
2. What other methods could be used to prove divisibility properties?
3. How does changing the modulus affect the induction proof?
4. Can you extend this result to other forms of exponential expressions?
5. How does the structure of prime factorization of 726 influence the solution?

**Tip**: When doing induction proofs, always verify the base case carefully. It anchors the entire argument.

Verifique por inducción que para todo n ≥ 1: 2 · 34^n - 3 · 23^n + 1 es divisible por 726.

Learn how to verify by mathematical induction that for all n ≥ 1, the expression 2 · 34^n - 3 · 23^n + 1 is divisible by 726. This problem explores divisibility, modular arithmetic, and exponentiation in high school-level math.

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