 # Important Questions for Class 11 Maths Chapter 4 - Principles of Mathematical Induction

Important questions for class 11 Maths Chapter 4 – Principles of Mathematical Induction are given here. Chapter 4 Mathematical Induction of class 11 includes problems or statements which involve mathematical relations. It is one of the important topics of class 11. Students can easily score marks in this chapter. We have given a few important questions in chapter 4 – Principles of Mathematical Induction on this page. Solving these problems will help you in scoring well in exams. Also, get all the important Maths questions from class 11 chapters at BYJU’S.

Class 11 Maths Chapter 4 – principles of Mathematical Induction cover the following important concepts such as introduction and motivation and the concept behind the principle of Mathematical induction.

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## Class 11 Chapter 4 – Principles of Mathematical Induction Important Questions with Solutions

Practice class 11 chapter 4 principles of Mathematical Induction problems given here.

Question 1:

Prove that 2n > n for all positive integers n by the Principle of Mathematical Induction

Solution:

Assume that P(n): 2n > n

If n =1, 21>1. Hence P(1) is true

Let us assume that P(k) is true for any positive integer k,

It means that, i.e.,

2k > 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. 2k > 2k

Now by using the property,

i.e., 2k+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) = n2 using the principle of mathematical induction.

Solution:

Given Statement: 1 + 3 + 5 + … + (2n – 1) = n2

Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n2 , for n ∈ N

Note that P(1) is true, since

P(1) : 1 = 12

Let P(k) is true for some k ∈ N,

It means that,

P(k) : 1 + 3 + 5 + … + (2k – 1) = k2

To prove that P(k + 1) is true, we have

1 + 3 + 5 + … + (2k – 1) + (2k + 1)

= k2 + (2k + 1)

= k2 + 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.

### Practice Problems for Class 11 Maths Chapter 4

Go through class 11 chapter 4 principles of Mathematical induction concepts and solve the practice problems

1. Show that n(n+1)(2n +1) is divisible by 6 for all n that belongs to N(Use the principle of mathematical induction).
2. Prove by induction that the sum of the cubes of three consecutive natural numbers is divisible by 9.
3. State the first principle of mathematical induction.
4. Prove by using the principle of mathematical induction: – 2 + 5 + 8 + 11 + …+(3n – 1) = 1/2n(3n+1).
5. 1 + 3 + 32 +… + 3n-1 = (3n – 1) / 2, Use mathematical induction to prove LHS = RHS.
6. Prove that number of subsets of a set containing n distinct elements is 2n, for all n ⊆ N.
7. If xn-1 is divisible by x-k, then the least positive integral value of k is:
a) 1
b) 2
c) 3
d) 4
8. Show that n5/5 + n3/3 + 7n/15 is a natural number for all n ⊆ N.