Test: Introduction To Mathematical Induction - Commerce MCQ

P(n) : (2n + 7) < (n + 3)² 
P(3) = [(2(3) + 7) < (3+ 3)²

P(n) = 4 + 8 + 12 +……+ 4n = 2n (n + 1)
P(6) = 4 + 8 + 12 +……+ 4(6) 
= 2(6)[(6) + 1)]

4^2x = 4^-6

2x = -6

So,  x = -3

Assume the inductive hypothesis that we can reach rung k. Then, we can reach rung k + 1.
Hence, P(k) → P(k + 1) is true for all positive integers k

Correct Answer :- b

Explanation : Let P(n) be the given statement, i.e., P(n) : (2n + 1) < 2n

for all natural

numbers, n ≥ 3. We observe that P(3) is true, since

2.3 + 1 = 7 < 8 = 23

Assume that P(n) is true for some natural number k, i.e., 2k + 1 < 2k

To prove P(k + 1) is true, we have to show that 2(k + 1) + 1 < 2k+1. Now, we have

2(k + 1) + 1 = 2 k + 3

= 2k + 1 + 2 < 2k + 2 < 2k

. 2 = 2k + 1

Thus P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers, n ≥ 3

To prove a statement by induction, we must prove parts 1) and 2) above. The hypothesis of Step 1) -- "The statement is true for n = k" -- is called the induction assumption, or the induction hypothesis. It is what we assume when we prove a theorem by induction.

The hypothesis of Step is a must for mathematical induction that is the statement is true for n = k, where n and k are any natural numbers, which is also called induction assumption or induction hypothesis.