Answer: (D) = 3
For n = 1, n3 + 2n = 1 + 2 = 3 which is divisible by 3 and none of the other given alternatives.
∴ We shall prove n3 + 2n divisible by 3 for all n∈N.
Let T(n) = n3 + 2n is divisible by 3.
Basic Step:
For n = 1, T(1) = n3 + 2n = 1 + 2 = 3 is divisible by 3 is true.
Induction Step:
Assume T(k) to be true, i.e., T(k) = k3 + 2k is divisible by 3
= k3 + 2k = 3m, where m∈N. ...(i)
Now we need to prove that T(k + 1) holds true, i.e.,
(k + 1)3 + 2(k + 1) is divisible by 3.
(k + 1)3 + 2(k + 1) = k3 + 3k2 + 3k +1 + 2k + 2
= (k3 + 2k) + (3k2 + 3k + 3)
= 3m + 3 (k2 + k + 1) (From (i))
⇒ T(k + 1) = (k + 1)3 + 2 (k + 1) is divisible by 3, whenever T(k) = k3 + 2k is divisible by 3.
⇒ n3 + 2n is divisible by 3 ∀ n∈N.