Prove that for all +ve integral values of n, 1 + 3 + 5 + .... + (2n �

1.	Prove that for all +ve integral values of n, 1 + 3 + 5 + .... + (2n – 1) = n2.
Answer» Let T(n) be the statement: 1 + 3 + 5 + ... + (2n – 1) = n² Basic Step: For n = 1, LHS = 1, RHS = 1² ⇒ LHS = RHS ⇒ T(1) is true Induction Step: Assume that T(k) is true, i.e., 1 + 3 + 5 + ... + (2k – 1) = k² To obtain T (k + 1), add the (k + 1)^th term = 2 (k + 1) – 1 = 2k + 2 – 1 = 2k + 1 to both the sides. Then, 1 + 3 + 5 + ... + (2k – 1) + (2k + 1) = k² + 2k + 1 ⇒ 1 + 3 + 5 + ... to (k + 1) terms = (k + 1)² Thus the statement is true for n = k + 1 under the assumption that statement is true for n = k Therefore, the statement 1 + 3 + 5 + ... to n terms = n² for every positive integer n.

Prove that for all +ve integral values of n, 1 + 3 + 5 + .... + (2n – 1) = n2.

Answer»

Let T(n) be the statement: 1 + 3 + 5 + ... + (2n – 1) = n²

Basic Step: For n = 1, LHS = 1, RHS = 1²

⇒ LHS = RHS ⇒ T(1) is true

Induction Step: Assume that T(k) is true, i.e., 1 + 3 + 5 + ... + (2k – 1) = k²

To obtain T (k + 1), add the (k + 1)^th term

= 2 (k + 1) – 1 = 2k + 2 – 1 = 2k + 1 to both the sides.

Then, 1 + 3 + 5 + ... + (2k – 1) + (2k + 1) = k² + 2k + 1

⇒ 1 + 3 + 5 + ... to (k + 1) terms = (k + 1)²

Thus the statement is true for n = k + 1 under the assumption that statement is true for n = k

Therefore, the statement 1 + 3 + 5 + ... to n terms = n² for every positive integer n.