For all natural number n, 2 + 4 + 6 + ..... + 2n equals(a) 2 (n + 1)

1.	For all natural number n, 2 + 4 + 6 + ..... + 2n equals(a) 2 (n + 1) (b) \(\frac{1}{2}\) n (n +2)(c) n (n + 1) (d) (n + 2) (n + 4)
Answer» Answer: (C) n(n+1) Let S_n = 2 + 4 + 6 + .... + 2n When n = 1, S_n = 2 Now from the options given, when n = 1, 2(n + 1) = 4, \(\frac{1}{2}n(n+2) =\frac{3}{2}\) ,n(n + 1) = 2, (n + 2) (n + 4) = 15 ∴ Sn ≠ 2(n + 1), Sn ≠ \(\frac{1}{2}\) n (n + 2), Sn ≠ (n + 2) (n + 4) for n = 1 Sn = n (n + 1) for n = 1 ∴ We need to prove 2 + 4 + 6 + .... + 2n = n (n + 1) ∀ n∈N. Let T(n) = 2 + 4 + 6 + .... + 2n = n (n + 1) Basic Step: For n =1, LHS = 2 × 1 = 2, RHS = 1 × (1 + 1) = 2 ⇒ LHS = RHS ⇒ T(1) is true. Induction Step: Assume T(k) is true, i.e., 2 + 4 + 6 + .... + 2k = k (k + 1) To obtain T(k + 1), we add the (k + 1)^th term, i.e, 2 (k + 1) to both the sides of T(k), i.e., 2 + 4 + 6 + ... + 2k + 2(k + 1) = k(k + 1) + 2(k + 1) = (k + 1) (k + 2) = (k + 1) ((k + 1) + 1) Thus the statement T(n) is true for n = k + 1, whenever it is true for n = k. Therefore by the principle of mathematical induction it is true for all n∈N

For all natural number n, 2 + 4 + 6 + ..... + 2n equals(a) 2 (n + 1) (b) \(\frac{1}{2}\) n (n +2)(c) n (n + 1) (d) (n + 2) (n + 4)

Answer»

Answer: (C) n(n+1)

Let S_n = 2 + 4 + 6 + .... + 2n

When n = 1, S_n = 2

Now from the options given, when n = 1,

2(n + 1) = 4, \(\frac{1}{2}n(n+2) =\frac{3}{2}\) ,n(n + 1) = 2, (n + 2) (n + 4) = 15

∴ Sn ≠ 2(n + 1), Sn ≠ \(\frac{1}{2}\) n (n + 2), Sn ≠ (n + 2) (n + 4) for n = 1

Sn = n (n + 1) for n = 1

∴ We need to prove 2 + 4 + 6 + .... + 2n = n (n + 1) ∀ n∈N.

Let T(n) = 2 + 4 + 6 + .... + 2n = n (n + 1)

Basic Step:

For n =1, LHS = 2 × 1 = 2, RHS = 1 × (1 + 1) = 2

⇒ LHS = RHS

⇒ T(1) is true.

Induction Step:

Assume T(k) is true, i.e.,

2 + 4 + 6 + .... + 2k = k (k + 1)

To obtain T(k + 1), we add the (k + 1)^th term, i.e, 2 (k + 1) to both the sides of T(k),

i.e., 2 + 4 + 6 + ... + 2k + 2(k + 1) = k(k + 1) + 2(k + 1)

= (k + 1) (k + 2) = (k + 1) ((k + 1) + 1)

Thus the statement T(n) is true for n = k + 1, whenever it is true for n = k.

Therefore by the principle of mathematical induction it is true for all n∈N