Let S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k2. Then, which of the

1.	Let S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k2. Then, which of the following is true?(a) S(1) is correct (b) S(k) ⇒ S (k + 1) (c) S(k) \(\neq\) S(k + 1) (d) Principle of mathematical induction can be used to prove the above formula.
Answer» Answer: (B) S(k) ⇒ S(k+1) S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k² Putting k = 1 on both the sides, we get LHS = 1, RHS = 3 + 1 = 4 ⇒ LHS ≠ RHS ⇒ S(1) is not true. Assume S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k² is true. Then, To find S(k + 1), add the (k + 1)^th term = (2 (k + 1) – 1) = 2k +1 on both the sides of S(k). ∴ S(k + 1) = 1 + 3 + 5 + .... + (2k – 1)+(2k + 1) = 3 + k² + 2k + 1 ⇒ 1 + 3 + 5 + .... + (2k – 1) + (2k + 1) = 3 + (k + 1)² ⇒ S(k + 1) is also true. ∴ S(k) ⇒ S(k + 1) is true

Let S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k2. Then, which of the following is true?(a) S(1) is correct (b) S(k) ⇒ S (k + 1) (c) S(k) \(\neq\) S(k + 1) (d) Principle of mathematical induction can be used to prove the above formula.

Answer»

Answer: (B) S(k) ⇒ S(k+1)

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

Putting k = 1 on both the sides, we get

LHS = 1, RHS = 3 + 1 = 4

⇒ LHS ≠ RHS

⇒ S(1) is not true.

Assume S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k² is true.

Then,

To find S(k + 1), add the (k + 1)^th term = (2 (k + 1) – 1) = 2k +1 on both the sides of S(k).

∴ S(k + 1) = 1 + 3 + 5 + .... + (2k – 1)+(2k + 1) = 3 + k² + 2k + 1

⇒ 1 + 3 + 5 + .... + (2k – 1) + (2k + 1) = 3 + (k + 1)²

⇒ S(k + 1) is also true.

∴ S(k) ⇒ S(k + 1) is true