The correct option is B n ∈ N
For n=1,LHS=4−1=3 and RHS=3So LHS=RHS⇒P(1) is true.
Let P(k) is true for some natural number k,i.e., 4k−1≥3k ...(1)We need to prove that P(k+1) is true whenever P(k) is true.Consider 4k+1−1=4.4k−1 =4(4k−1+1)−1 =4.(4k−1)+4−1 ≥4.3k+3 (Using (1)] >4.3k >3.3k=3k+1i.e., 4k+1−1≥3k+1Therefore, P(k+1) is true whenever P(k) is true.Hence, by principal of mathematical induction, P(n) is true for every natural number.