To prove that $f(n) = 2^{n+2} + 3^{2n+1}$ is divisible by 7 for all positive integers $n$, we will use mathematical induction.

Step 1: Base Case

We need to check if the statement holds for $n = 1$.

For $n = 1$, $f(1) = 2^{1+2} + 3^{2(1) + 1} = 2^3 + 3^3 = 8 + 27 = 35.$ Clearly, $35$ is divisible by 7, since $35 \div 7 = 5$.

Thus, the base case is true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary positive integer $k$, i.e., assume: $f(k) = 2^{k+2} + 3^{2k+1}$ is divisible by 7. This means: $2^{k+2} + 3^{2k+1} \equiv 0 \pmod{7}.$

Step 3: Inductive Step

We need to prove that the statement holds for $k+1$, i.e., we need to show: $f(k+1) = 2^{(k+1)+2} + 3^{2(k+1)+1} = 2^{k+3} + 3^{2k+3}$ is divisible by 7.

We begin by expressing $f(k+1)$: $f(k+1) = 2^{k+3} + 3^{2k+3}.$ We can rewrite this as: $f(k+1) = 2^{k+2} \cdot 2 + 3^{2k+1} \cdot 3^2.$ From our inductive hypothesis, we know that: $2^{k+2} + 3^{2k+1} \equiv 0 \pmod{7}.$ Now, we need to consider $f(k+1) \mod 7$ in more detail.

First, note that: $2^3 = 8 \equiv 1 \pmod{7} \quad \text{and} \quad 3^2 = 9 \equiv 2 \pmod{7}.$ Thus, the expression for $f(k+1)$ modulo 7 becomes: $f(k+1) = 2 \cdot (2^{k+2}) + 2 \cdot (3^{2k+1}).$ Since ( 2^{k+2} + 3^{2k+1}\equiv 0\