Let P(n) = 3^{2n+1} + 2^{n+2} where n in N. T | vedclass

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(B) Given $P(n) = 3^{2n+1} + 2^{n+2}$.For $n = 1$, $P(1) = 3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35$.Since $35 = 5 	imes 7$, $P(1)$ is divisibl

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there exists a prime integer which divides $P(n)$.

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(B) Given $P(n) = 3^{2n+1} + 2^{n+2}$.For $n = 1$, $P(1) = 3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35$.Since $35 = 5 	imes 7$, $P(1)$ is divisible by the prime integers $5$ and $7$.Thus, there exists at least one prime integer that divides $P(n)$ for $n = 1$.For any $n \in N$, $P(n) > 1$, and by the Fundamental Theorem of Arithmetic, every integer greater than $1$ has at least one prime divisor.Therefore, there exists a prime integer which divides $P(n)$.

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

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