Right answer is (a) 3
To elaborate: P(n) = n^3 + 5n
P(1) = 1 + 5
P(1) = 6
We assume the P(k) is true and divisible by 6.
P(k) = k^3 + 5k is divisible by 6 and can be written as 6c or 3 x 2c
We need to prove that P(k + 1) is divisible by 6
P(k + 1) = (k + 1)^3 + 5(k + 1)
P(k + 1) = k^3 + 1 + 3k^2 + 3k + 5k + 5
P(k + 1) = (k^3 + 5k) + 3k^2 + 3k + 6
P(k + 1) = 6c + 3(k^2 + k + 2)
P(k + 1) = (3 x 2c) + 3(k^2 + k + 2)
Therefore, P(k + 1) is definitely divisible by 3