For all `n in N, 3.5^(2n+1) + 2^(3n+1)` is divisible by:
(i) 17
(ii) 19
(iii) 23
(iv) 25

Options

A. 17
B. 19
C. 23
D. 25

Answer

To solve the problem, we need to prove that the expression \( P(n) = 3 \cdot 5^{2n+1} + 2^{3n+1} \) is divisible by certain numbers for all natural numbers \( n \).

### Step 1: Base Case
We start by checking the base case \( n = 1 \).

\[ P(1) = 3 \cdot 5^{2 \cdot 1 + 1} + 2^{3 \cdot 1 + 1} \]
\[ = 3 \cdot 5^{3} + 2^{4} \]
\[ = 3 \cdot 125 + 16 \]
\[ = 375 + 16 = 391 \]

Now we check if 391 is divisible by 17, 19, 23, and 25.

- **Divisibility by 17**:
\[ 391 \div 17 = 23 \quad \text{(exact)} \]
- **Divisibility by 19**:
\[ 391 \div 19 \approx 20.5789 \quad \text{(not exact)} \]
- **Divisibility by 23**:
\[ 391 \div 23 = 17 \quad \text{(exact)} \]
- **Divisibility by 25**:
\[ 391 \div 25 = 15.64 \quad \text{(not exact)} \]

Thus, for \( n = 1 \), \( P(1) \) is divisible by 17 and 23.

### Step 2: Inductive Step
Next, we will assume that \( P(k) \) is divisible by 17 and 23 for some \( k \). We need to show that \( P(k+1) \) is also divisible by these numbers.

\[ P(k+1) = 3 \cdot 5^{2(k+1)+1} + 2^{3(k+1)+1} \]
\[ = 3 \cdot 5^{2k+3} + 2^{3k+4} \]
\[ = 3 \cdot 5^{2k+1} \cdot 5^2 + 2^{3k+1} \cdot 2^3 \]
\[ = 3 \cdot 5^{2k+1} \cdot 25 + 2^{3k+1} \cdot 8 \]

Now, we can express \( P(k) \) in terms of \( P(k+1) \):
\[ P(k+1) = 25 \cdot P(k) + 8 \cdot 2^{3k+1} \]

Since we assumed \( P(k) \) is divisible by 17 and 23, \( 25 \cdot P(k) \) is also divisible by 17 and 23.

Now, we need to check \( 8 \cdot 2^{3k+1} \):
- \( 2^{3k+1} \) is clearly divisible by 2, and since \( 8 \) is \( 2^3 \), \( 8 \cdot 2^{3k+1} \) is divisible by \( 2^{3+1} = 16 \).

Thus, both terms in \( P(k+1) \) are divisible by 17 and 23.

### Conclusion
By the principle of mathematical induction, \( P(n) = 3 \cdot 5^{2n+1} + 2^{3n+1} \) is divisible by 17 and 23 for all \( n \in \mathbb{N} \).

### Final Answer
The expression is divisible by:
- (i) 17
- (iii) 23