Right answer is (d) m^3 + 3m
Best explanation: The required answer is, m^3 + 3m. Now, by induction hypothesis, we have to prove for m=k, k^3+3k is divisible by 4. So, (k + 1)^3 + 3 (k + 1) = k^3 + 3 k^2 + 6 k + 4
= [k^3 + 3 k] + [3 k^2 + 3 k + 4] = 4M + (12k^2 + 12k) – (8k^2 + 8k – 4), both the terms are divisible by 4. Hence (k + 1)^3 + 3 (k + 1) is also divisible by 4 and hence it is proved for any integer m.