Answer: (B) an integral multiple of 6
When n = 1, n(n+1)(2n+1) = (1) (2) (3) = 6, which is an integral multiple of 6.
It is neither an odd integer nor a perfect square.
Using the principle of mathematical induction, we shall now show that the expression n(n + 1)(2n + 1) is an integral multiple of 6 ∀ n ∈ N.
Assume T(n) = n(n+1)(2n+1) = 6x where x∈N.
Basic Step:
T(1) is true as shown above.
Induction Step:
Let T(k) be true for all k∈N.
⇒ k (k +1) (2k +1) = 6x, where x∈N. .....(i)
For T(k + 1), we replace k by (k + 1) in the given expression, i.e.,
T(k + 1) = (k + 1) (k + 2) (2 (k + 1) + 1)
= (k + 1) (k + 2) ((2k + 1) + 2)
= (k + 1) (k + 2) (2k + 1) + 2 (k + 1) (k + 2)
= k (k + 1) (2k + 1) + 2 (k + 1) (2k + 1) + 2 (k + 1) (k + 2)
= k (k + 1) (2k + 1) + 2 (k + 1) [(2k + 1) + (k + 2)]
= 6x + 2(k + 1)(3k + 3)
= 6x + 6(k + 1)2
= 6 (x + (k + 1)2) = 6 × a positive integer
∴ T(k) is true ⇒ T(k + 1) is true.
∴ T(n) is true for all n∈N