Important Questions for Class 11 Maths Chapter 4: Class 11 Maths Chapter 4 Principle of Mathematical Induction is a foundational topic in higher mathematics that strengthens students logical thinking and proof skills.
Important questions in this chapter focus on applying the principle of mathematical induction to prove statements about sequences, inequalities, divisibility and identities. These questions help students practice structuring mathematical proofs step-by-step, making sure to establish a base case and then proving that if the statement holds for one integer, it also holds for the next. Mastery of this chapter is important, as it sets the stage for more advanced proof techniques used in calculus, algebra and beyond. Practicing important questions from Chapter 4 helps students gain confidence in their reasoning skills and prepares them for a range of mathematical problems in competitive exams and further studies.Important Questions for Class 11 Maths Chapter 4 PDF
Question 1:
Prove that 2 n > n for all positive integers n by the Principle of Mathematical InductionSolution:
Assume that P(n): 2 n > n If n =1, 2 1 >1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 2 k > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by 2, we get 2. 2 k > 2k Now by using the property, i.e., 2 k+1 > 2k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.Question 2:
Prove that 1 + 3 + 5 + … + (2n – 1) = n 2 using the principle of mathematical induction.Solution:
Given Statement: 1 + 3 + 5 + … + (2n – 1) = n 2 Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n 2 , for n ∈ N Note that P(1) is true, since P(1) : 1 = 1 2 Let P(k) is true for some k ∈ N, It means that, P(k) : 1 + 3 + 5 + … + (2k – 1) = k 2 To prove that P(k + 1) is true, we have 1 + 3 + 5 + … + (2k – 1) + (2k + 1) = k 2 + (2k + 1) = k 2 + 2k + 1 By using the formula, the above form can be written as: = (k + 1) 2 Hence, P(k + 1) is true, whenever P(k) is true. Therefore, P(n) is true for all n ∈ N is proved by the principle of mathematical induction.Question 3:
Show that 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n by the Principle of Mathematical Induction.Solution:
Assume that P(n) be the given statement, that is P(n) : 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n. It is noted that P (1) is true, since P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1. Let P(n) is true for some natural number k, It means that P(k) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! = (k + 1)! – 1 Inorder to prove P (k + 1) is true, we have P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! + (k + 1) × (k + 1)! = (k + 1)! – 1 + (k + 1)! × (k + 1) Now, simplify the above form, we get = (k + 1 + 1) (k + 1)! – 1 = (k + 2) (k + 1)! – 1 = ((k + 2)! – 1 Therefore, P (k + 1) is true, whenever P (k) is true. Hence, P(n) is true for all natural number n is proved using the Principle of Mathematical Induction.