If ‘n’ be any positive integer, then n (n + 1) (2n + 1) is(a) an o

1.	If ‘n’ be any positive integer, then n (n + 1) (2n + 1) is(a) an odd integer (b) an integral multiple of 6 (c) a perfect square (d) None of these
Answer» 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

If ‘n’ be any positive integer, then n (n + 1) (2n + 1) is(a) an odd integer (b) an integral multiple of 6 (c) a perfect square (d) None of these

Answer»

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