Question

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 1

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