Prove by induction that f(n) = 2^(k + 2) + 3^(3k + 1) is divisible by 7 for all positive n.

First we establish our base case: f(0) = 2² + 3¹ = 4 + 3 = 7, so clearly f(0) is divisible by 7.Now. by the inductive hypothesis. we assume that f(k) is divisible by 7, and attempt to show that this implies f(k+1) is also divisible by 7.f(k + 1) = 2^{k + 3} + 3^{2(k + 1) + 1} = 2^{k + 3} + 3^{2k + 3} = 2 * 2^{k + 2} + 9 * 3^{2k + 1} So f(k + 1) mod 7 === 2 * 2^{k + 2} + 2 * 3^{2k + 1} (since 9 mod 7 === 2). So f(k + 1) mod 7 === 2 * (2^{k + 2} + 3^{2k + 1}) = 2 * f(k).These for f(k + 1) mod 7 === 0, hence f(k + 1) is divisible by 7 if f(k) is divisible by 7, hence f(n) is divisibile by 7 for all n >= 0.