JEE Advance - Mathematics (1982 - No. 23)
Prove that $${7^{2n}} + \left( {{2^{3n - 3}}} \right)\left( {3n - 1} \right)$$ is divisible by 25 for any natural number $$n$$.
The statement is only true for even values of n.
The statement is false for all natural numbers n.
The statement is true for all natural numbers n. Use induction with base case n=1 and showing divisibility by 25 for the inductive step.
The statement is true only for n < 5.
The statement is true for all natural numbers n. Base case n=1, assuming true for n=k implies true for n=k+1. Express 7^(2(k+1)) + 2^(3(k+1)-3)(3(k+1)-1) = 49 * 7^(2k) + 8 * 2^(3k-3) (3k+2) = 49[7^(2k) + 2^(3k-3)(3k-1)] + 2^(3k-3) [8(3k+2) - 49(3k-1)] = 49[7^(2k) + 2^(3k-3)(3k-1)] + 2^(3k-3) [-123k + 65]. -123k + 65 = -125k + 2k + 50 + 15 = 25(-5k + 2) + (2k+15). Needs additional proof that 2k+15 is divisible by 25.