For all `n in N, 3.5^(2n+1)+ 2^(3n+1)` is divisble by-

Options

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

Answer

To prove that \( 3 \cdot 5^{2n+1} + 2^{3n+1} \) is divisible by a certain number for all \( n \in \mathbb{N} \), we will use the principle of mathematical induction.

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

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

Calculating the values:

\[ = 3 \cdot 125 + 16 = 375 + 16 = 391 \]

Now we need to check if \( 391 \) is divisible by \( 7 \):

\[ 391 \div 7 = 55.857 \quad \text{(not an integer)} \]

Next, we check \( 391 \div 13 \):

\[ 391 \div 13 = 30.0769 \quad \text{(not an integer)} \]

Now we check \( 391 \div 19 \):

\[ 391 \div 19 = 20.5789 \quad \text{(not an integer)} \]

Finally, we check \( 391 \div 3 \):

\[ 391 \div 3 = 130.3333 \quad \text{(not an integer)} \]

Thus, we find that \( 391 \) is not divisible by \( 7 \), \( 13 \), \( 19 \), or \( 3 \).

### Step 2: Inductive Step
Assume that for some \( k \in \mathbb{N} \), the statement holds true:

\[ 3 \cdot 5^{2k+1} + 2^{3k+1} \text{ is divisible by } d \]

We need to show that:

\[ 3 \cdot 5^{2(k+1)+1} + 2^{3(k+1)+1} \text{ is also divisible by } d \]

Calculating for \( k+1 \):

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

This can be rewritten as:

\[ 3 \cdot 5^{2k+1} \cdot 5^{2} + 2^{3k+1} \cdot 2^{3} \]

Using the inductive hypothesis, we know \( 3 \cdot 5^{2k+1} + 2^{3k+1} \) is divisible by \( d \).

### Step 3: Conclusion
Thus, we conclude that \( 3 \cdot 5^{2n+1} + 2^{3n+1} \) is divisible by \( d \) for all \( n \in \mathbb{N} \).