Question

Let \(P(n)=3^{2n+1}+2^{n+2}\) where n∈N. Then

• A. P(n) is not divisible by any prime integer
• B. There exists a prime integer which divides \(P(n)\).
• C. P(n) is divisible by 5 for all n∈N
• D. P(n) is divisible by 3 for all n∈N

Answer 1

The Correct Option is B

Step 1: Define the function

$P(n) = 3^{2n+1} + 2^{n+2}$ 

Step 2: Try small values

$P(1) = 3^3 + 2^3 = 27 + 8 = 35 = 5 \cdot 7$

$P(2) = 3^5 + 2^4 = 243 + 16 = 259 = 7 \cdot 37$

$P(3) = 3^7 + 2^5 = 2187 + 32 = 2219$, not divisible by 5 or 3

Step 3: General form and induction for divisibility by 7

Assume $P(k) = 3^{2k+1} + 2^{k+2}$ is divisible by 7

Then: $P(k+1) = 3^{2k+3} + 2^{k+3} = 9 \cdot 3^{2k+1} + 2 \cdot 2^{k+2}$

Substitute: $P(k+1) = 9(7m - 2^{k+2}) + 2 \cdot 2^{k+2} = 63m - 7 \cdot 2^{k+2}$

$= 7(9m - 2^{k+2})$ ⟹ divisible by 7

Conclusion: There exists a prime integer (7) which divides $P(n)$ for all $n \in \mathbb{N}$.

Correct answer: There exists a prime integer which divides $P(n)$