For all `n in N, 2^(n+1) + 3^(2n-1)` is divisible by:
(i) 5
(ii) 7
(iii) 14
(iv) 135

Options

A. 5
B. 7
C. 14
D. 135

Answer

To determine if the expression \( 2^{n+1} + 3^{2n-1} \) is divisible by the numbers 5, 7, 14, and 135 for all \( n \in \mathbb{N} \), we will evaluate the expression for different values of \( n \) and check the divisibility.

### Step 1: Evaluate for \( n = 1 \)

For \( n = 1 \):
\[ 2^{1+1} + 3^{2 \cdot 1 - 1} = 2^2 + 3^1 = 4 + 3 = 7 \]
**Divisibility Check:**
- \( 7 \div 5 \) gives a remainder of 2 (not divisible by 5)
- \( 7 \div 7 = 1 \) (divisible by 7)
- \( 7 \div 14 \) gives a remainder of 7 (not divisible by 14)
- \( 7 \div 135 \) gives a remainder of 7 (not divisible by 135)

### Step 2: Evaluate for \( n = 2 \)

For \( n = 2 \):
\[ 2^{2+1} + 3^{2 \cdot 2 - 1} = 2^3 + 3^3 = 8 + 27 = 35 \]
**Divisibility Check:**
- \( 35 \div 5 = 7 \) (divisible by 5)
- \( 35 \div 7 = 5 \) (divisible by 7)
- \( 35 \div 14 \) gives a remainder of 7 (not divisible by 14)
- \( 35 \div 135 \) gives a remainder of 35 (not divisible by 135)

### Step 3: Evaluate for \( n = 3 \)

For \( n = 3 \):
\[ 2^{3+1} + 3^{2 \cdot 3 - 1} = 2^4 + 3^5 = 16 + 243 = 259 \]
**Divisibility Check:**
- \( 259 \div 5 \) gives a remainder of 4 (not divisible by 5)
- \( 259 \div 7 \) gives a remainder of 0 (divisible by 7)
- \( 259 \div 14 \) gives a remainder of 5 (not divisible by 14)
- \( 259 \div 135 \) gives a remainder of 124 (not divisible by 135)

### Conclusion

From the evaluations:
- The expression \( 2^{n+1} + 3^{2n-1} \) is divisible by \( 7 \) for \( n = 1, 2, 3 \).
- It is not consistently divisible by \( 5, 14, \) or \( 135 \).

Thus, the expression is divisible by **7** for all \( n \in \mathbb{N} \).