If n is a positive integer, then `2.4^(2n + 1) + 3^(3n+1) ` is divisible by :

Answer

To determine the divisibility of the expression \( 2 \cdot 4^{2n + 1} + 3^{3n + 1} \) for positive integers \( n \), we will analyze the expression step by step.

### Step 1: Rewrite the expression
The expression is given as:
\[ 2 \cdot 4^{2n + 1} + 3^{3n + 1} \]
We can rewrite \( 4^{2n + 1} \) as \( (2^2)^{2n + 1} = 2^{4n + 2} \). Thus, the expression becomes:
\[ 2 \cdot 2^{4n + 2} + 3^{3n + 1} \]
This simplifies to:
\[ 2^{4n + 3} + 3^{3n + 1} \]

### Step 2: Substitute values for \( n \)
To find a pattern, we will substitute small positive integer values for \( n \).

#### For \( n = 1 \):
\[ 2^{4 \cdot 1 + 3} + 3^{3 \cdot 1 + 1} = 2^{7} + 3^{4} = 128 + 81 = 209 \]

#### For \( n = 2 \):
\[ 2^{4 \cdot 2 + 3} + 3^{3 \cdot 2 + 1} = 2^{11} + 3^{7} = 2048 + 2187 = 4235 \]

#### For \( n = 3 \):
\[ 2^{4 \cdot 3 + 3} + 3^{3 \cdot 3 + 1} = 2^{15} + 3^{10} = 32768 + 59049 = 91817 \]

### Step 3: Check divisibility by 11
Now, we will check the results for divisibility by 11.

- For \( n = 1 \):
\[ 209 \div 11 = 19 \quad \text{(remainder 0, divisible)} \]

- For \( n = 2 \):
\[ 4235 \div 11 = 385 \quad \text{(remainder 0, divisible)} \]

- For \( n = 3 \):
\[ 91817 \div 11 = 8347 \quad \text{(remainder 0, divisible)} \]

### Conclusion
From the calculations, we observe that the expression \( 2^{4n + 3} + 3^{3n + 1} \) is divisible by 11 for the values of \( n \) tested. We can conclude that for any positive integer \( n \), the expression is divisible by 11.

Thus, the final answer is:
\[ \text{The expression } 2 \cdot 4^{2n + 1} + 3^{3n + 1} \text{ is divisible by } 11. \]