Use induction to show that 5^(n+1)−4n+11 is divisible by 16 for all n≥1.

We are asked to prove, by mathematical induction, that \( 5^{n+1} - 4n + 11 \) is divisible by 16 for all \( n \geq 1 \). This means we need to show that \( 5^{n+1} - 4n + 11 \) is divisible by 16 for every integer \( n \geq 1 \), i.e.,

\[
5^{n+1} - 4n + 11 \equiv 0 \pmod{16}
\]

### Step 1: Base Case
We first check the base case where \( n = 1 \).

For \( n = 1 \), the expression becomes:
\[
5^{1+1} - 4(1) + 11 = 5^2 - 4 + 11 = 25 - 4 + 11 = 32
\]
Since \( 32 \) is divisible by 16, the base case holds.

### Step 2: Inductive Hypothesis
Now, assume that the statement holds for some arbitrary \( n = k \), i.e., assume that:
\[
5^{k+1} - 4k + 11 \equiv 0 \pmod{16}
\]
This means:
\[
5^{k+1} - 4k + 11 = 16m \quad \text{for some integer } m.
\]

### Step 3: Inductive Step
We need to show that the statement holds for \( n = k+1 \), i.e., we need to prove that:
\[
5^{(k+1)+1} - 4(k+1) + 11 \equiv 0 \pmod{16}.
\]
Simplifying the expression:
\[
5^{(k+1)+1} - 4(k+1) + 11 = 5^{k+2} - 4(k+1) + 11 = 5 \cdot 5^{k+1} - 4k - 4 + 11.
\]
We can group the terms as follows:
\[
5 \cdot 5^{k+1} - 4k - 4 + 11 = 5 \cdot (5^{k+1} - 4k + 11) - 4.
\]

By the inductive hypothesis, we know that \( 5^{k+1} - 4k + 11 \equiv 0 \pmod{16} \), so we can replace \( 5^{k+1} - 4k + 11 \) with \( 16m \) for some integer \( m \):
\[
5 \cdot (16m) - 4 = 80m - 4.
\]

Clearly, \( 80m \) is divisible by 16, and \( -4 \equiv 12 \pmod{16} \), so:
\[
80m - 4 \equiv 0 \pmod{16}.
\]

Thus, the expression \( 5^{k+2} - 4(k+1) + 11 \equiv 0 \pmod{16} \), completing the inductive step.

### Conclusion:
Since both the base case and the inductive step have been proven, by mathematical induction, \( 5^{n+1} - 4n + 11 \) is divisible by 16 for all \( n \geq 1 \).

Would you like any more details or have further questions?

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Here are five related questions that extend the idea:
1. How does modular arithmetic simplify proofs in number theory?
2. Can we generalize the inductive proof method for expressions with higher powers of numbers?
3. How can mathematical induction be used to prove divisibility properties involving multiple variables?
4. How does the inductive hypothesis relate to the structure of recursive sequences?
5. What are the limitations of mathematical induction in certain types of number theory problems?

**Tip**: When using induction, always clearly distinguish between the base case, inductive hypothesis, and inductive step to ensure a structured approach.

Explore how to use mathematical induction to prove that the expression 5^(n+1) - 4n + 11 is divisible by 16 for all n ≥ 1. This detailed guide covers modular arithmetic, number theory, and the principle of mathematical induction. Suitable for advanced high school and college-level students.

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