By principle of mathematical induction, \(\forall \mathrm{n} \in \mathrm{N}\), \(5^{2 \mathrm{n}+1}+3^{\mathrm{n}+2} \cdot 2^{\mathrm{n}-1}\) is divisible by (a) 19 (b) 18 (c) 17 (d) 14

From the given expression, when n = 1: \[5^{2(1) + 1} + 3^{1 + 2} \cdot 2^{1 - 1} = 5^3 + 3^3 \cdot 2^0 = 125 + 27 = 152 \] Now we test this value for divisibility: (a) $$152 \pmod{19} = 0$$ Hence divisible by 19 (b) $$152 \pmod{18} = 16$$ Not divisible by 18 (c) $$152 \pmod{17} = 1$$ Not divisible by 17 (d) $$152 \pmod{14} = 4$$ Not divisible by 14 We see that the expression is divisible by 19 for n=1.

Now assume that for n=k, the expression is divisible by 19: \[ 5^{2k + 1} + 3^{k + 2} \cdot 2^{k - 1} \equiv 0 \pmod{19} \tag{1} \]

Now, we need to prove that the expression is also divisible by 19, for n=k+1: \[ 5^{2(k+1) + 1} + 3^{(k+1) + 2} \cdot 2^{(k+1) - 1} \] Expanding and simplifying the expression, we get: \[ 5^{2k + 3} + 3^{k + 3} \cdot 2^k \] We can write the expression as: \[ 5^2 \cdot 5^{2k + 1} + 3 \cdot 3^{k + 2} \cdot 2^k \] Now, using the assumption in (1), we can rewrite the expression as follows: \[ 5^2 (19m - 3^{k + 2} \cdot 2^{k - 1}) + 3 \cdot 3^{k + 2} \cdot 2^k \] \[ \Rightarrow 19(25m) - 5^2 \cdot 3^{k + 2} \cdot 2^{k - 1} + 3 \cdot 3^{k + 2} \cdot 2^k \] \[ \Rightarrow 19(25m) + 3^{k + 2} \cdot 2^k - 5^2 \cdot 3^{k + 2} \cdot 2^{k - 1} \] Now, factoring by a common term, we can rewrite the expression as: \[ 19(25m) + 3^{k + 2} \cdot 2^k (1 - 5^2 \cdot 2^{-1}) \] As all terms are divisible by 19, this expression is also divisible by 19. So, we have proved the divisibility for n=k+1 using our assumption for n=k. By the Principle of Mathematical Induction, the expression is divisible by 19 for all natural numbers n. Therefore, the answer is (a) 19.