For each \(n \in \mathrm{N}, 2 \cdot 4^{2 \mathrm{n}+1}+3^{3 \mathrm{n}+1}\) is divisible by (a) 2 (b) 9 (c) 3 (d) 11

First, let's rewrite the expression more simply: Given \(P(n) = 2\cdot 4^{2n+1} + 3^{3n+1}\) We notice that the product \(4^{2n+1}\) can be rewritten as \(2^{2(2n+1)}\), so: \(P(n) = 2\cdot (2^{2(2n+1)}) + 3^{3n+1}\)

Now we will evaluate the expression P(n) modulo 2, 3, 9, and 11 to see if any required divisibility holds for all natural numbers n. 1. Modulo 2: \(P(n) \equiv 0 \pmod{2}\) as the first term is always \(2\cdot(2^{2(2n+1)})\), and that will always have a factor of 2. So, P(n) is divisible by 2 for all n. 2. Modulo 3: \(P(n) \equiv 2\cdot (2^{2(2n+1)}) + 3^{3n+1} \pmod{3}\) Notice that the first term has only factors of 2, so it will never have a factor of 3. On the other hand, the second term has a factor of 3. Thus, P(n) can be divisible by 3 only if the first term is congruent to -1 modulo 3. \(2\cdot (2^{2(2n+1)}) \equiv -1 \pmod{3}\) However, since the first term is always even, this condition can never hold since -1 is odd, and an even number can never be congruent to an odd number. Therefore, P(n) cannot be divisible by 3. 3. Modulo 9: \(P(n) \equiv 2\cdot (2^{2(2n+1)}) +3^{3n+1} \pmod{9}\), let's compute the first few values of \(P(n)\) modulo 9 and see if we find a pattern: \(P(0) = 36 \equiv 0 \pmod{9}\) \(P(1) = 16408 \equiv 3 \pmod{9}\) \(P(2) = 68719476762 \equiv 3 \pmod{9}\) As we can see, the pattern does not hold consistently; P(n) is not always divisible by 9. 4. Modulo 11: \(P(n) \equiv 2\cdot (2^{2(2n+1)}) +3^{3n+1} \pmod{11}\), let's compute the first few values of \(P(n)\) modulo 11 and see if we find a pattern: \(P(0) = 36 \equiv 3 \pmod{11}\) \(P(1) = 16408 \equiv 9 \pmod{11}\) \(P(2) = 68719476762 \equiv 8 \pmod{11}\) As we can see, the pattern does not hold consistently; P(n) is not always divisible by 11. So, the only valid option among the given choices is: (a) 2