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