1.

Using the method of induction, show that 1 + 2 + 3 + ... + n = \(\frac{1}{2}\) n (n + 1), for all n∈N

Answer»

Let T (n) = 1 + 2 + 3 + ... + n = \(\frac{1}{2}\) n (n + 1) 

Basic Step: For n = 1, 

LHS = T (1) = 1,   RHS = \(\frac{1}{2}\) × 1 × 2 = 1 ⇒ LHS = RHS ⇒ T(1) is true. 

Induction Step: Assume that T (k) is true, i.e., 

1 + 2 + 3 + ... + k = \(\frac{1}{2}\) k (k + 1) 

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

1 + 2 + 3 + ... + k + (k + 1) = \(\frac{1}{2}\) k (k + 1) + (k + 1) 

⇒ 1 + 2 + 3 + ... + k + (k + 1) = (k + 1) \(\big(\frac{k}{2}+1\big)\)

⇒ 1 + 2 + 3 + ... + k + (k + 1) = \(\frac{1}{2}\) (k + 1) (k + 2) 

⇒ Thus the statement T(n) is true for n = k + 1 under the assumption that T(k) is true. Therefore, by the principle of mathematical induction, the statement is true for every +ve integer n.


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