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Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:43 PM IST

**NCERT Exemplar Class 11 Maths solutions chapter 4** Principle of Mathematical Induction deals with various problems and finds its relevance through the method of deductive reasoning which has been considered as an important principle in mathematics for ages. NCERT Exemplar Class 11 Maths chapter 4 solutions covers the application of both deductive and inductive reasoning for proving the applicability of a generalized principle to any specific case. NCERT Exemplar solutions for Class 11 Maths chapter 4 also describes how mathematical induction can be used effectively to prove different mathematical statements. The principle of mathematical deduction was gradually developed over the years to help with the formulation and proving of various statements of science as well as mathematics.

This Story also Contains

- NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction
- Important Notes from NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction
- Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 4
- NCERT Solutions for Class 11 Mathematics Chapters

Question:1

Give an example of a statement which it true for all but are not true. Justify your answer.

Answer:

Given:

which it true for all but are not true.

Let us consider P (n) be

Examples of the given statements are-

Question:2

Give an example of a statement P (n) which is true for all n. Justify your answer.

Answer:

P (n) is true for all n …….. (Given)

Now, let P (n) be,

is true for all k

Thus, P (n) is true for all n.

Question:3

Answer:

P (n) = is divisible by 3 …….. (Given)

Now, we’ll substitute different values for n,

, viz. divisible by 3

, divisible by 3.

, divisible by 3

Now, let , is divisible by 3

Thus,

We also get that,

, is divisible by 3

Thus, P (k+1) is also true,

Hence, by mathematical induction,

For each natural no. n it is true that, is divisible by 3

Question:4

is divisible by7, for all natural numbers n.

Answer:

Now, we'll substitute different values for n,

Now, let us consider,

We also get that,

Hence, by mathematical induction, For each natural no. n it is true that,

Question:5

is divisible by 3, for all natural numbers n.

Answer:

is divisible by 3... given

Now, we’ll substitute different values for n,

, is divisible by 3

, is divisible by 3

, is divisible by 3

Now, let us consider,

be divisible by 3

Thus,

We also get that,

= is divisible by 3

Thus, P (k+1) is also true,

Hence, by mathematical induction,

For each natural no. n it is true that, is divisible by 3.

Question:6

is divisible by 8, for all natural numbers n.

Answer:

**p(n)= is divisible by 8.......(given)**

Now, we’ll substitute different values for n,

, is divisible by 8

, is divisible by 8

, is divisible by 8

Now, let us consider,

be divisible by 8

Thus, , is divisible by 8

Hence, by mathematical induction,

For each natural no. n it is true that, is divisible by 8.

Question:7

For any natural number n, is divisible by 5.

Answer:

P(n)= is divisible by 5. ...............(given)

Now, we will substitute different values for n,

, is divisible by 5

, is divisible by 5

, is divisible by 5

, is divisible by 5

Now, let us consider that,

be divisible by 5.

Thus,

We also get,

, is divisible by 5

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, is divisible by 5.

Question:8

For any natural number n, is divisible by x -y, where x integers with.

Answer:

P(n)= is divisible by x -y, where x integers with. ........(given)

Now, we’ll substitute different values for n,

, is divisible by x-y

is divisible by x-y

, is divisible by x-y

Now, let us consider,

, is divisible by x-y

Thus,

We also get that

, is divisible by x-y

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, is divisible by x-y, x integers with x*≠*y

Question:9

** is divisible by 6, for each natural number n. **

Answer:

**P(n)= is divisible by 6**

Now, we’ll substitute different values for n,

, is divisible by 6

Now, let us consider,

be divisible by 6

Thus,

We also get that,

,

is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, **P(n)= is divisible by 6.**

Question:10

is divisible by 6, for each natural number n.

Answer:

**P(n)= is divisible by 6, for each natural number n.**

Now, we’ll substitute different values for n,

, is divisible by 6

, is divisible by 6

Let us consider,

be divisible by 6

Thus,

We also get that,

, is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, **P(n)= is divisible by 6.**

Question:11

Answer:

P(n)= for .....(given)

Let us consider

to be true,

Thus,

Now, since, for

Thus,

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n^{2}< 2^{n} it is true that, P(n)= for ..

Question:12

Answer:

.....(given)

Now, we will substitute different values for n,

Now, let us consider,

To be true,

Thus,

Now, we know that,

But also,

Thus,

Thus, P(k+1) is true.

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) =

Question:13

Answer:

P(n) is ; .

Thus,

Hence, by mathematical induction,

For each natural no. ,P(n) is

Question:14

Answer:P(n)=** **

Now, we will substitute different values for n,

, is true

, is true

, is true

Now, let us consider,

to be true,

Thus,

P(k+1) is

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)=

Question:15

Answer:

P(n) is ** ..................given**

Now, we’ll substitute different values for n,

, is true

, is true

, is true

Now, let us consider,

to be true

Thus,

P(k+1) is

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) is **.**

Question:16

Answer:

P(n)=

Now, we’ll substitute different values for n,

P(1) = 1

= 1(2-1), is true

P(2) = 1+5 = 6

= 2(4-1), is true

Now, let us consider,

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)=

Question:17

A sequence is defined by letting for all natural numbers k ≥ 2. Show that for all natural numbers.

Answer:

Given

Thus, a_{n+1} is true if a_{n} is true

Hence, by mathematical induction,

For each natural no. n it is true that,

Question:18

Answer:

Given:

The sequence b_{0}, b_{1}, b_{2},... if defined if we let, & for all natural numbers k. Thus,

Now, let us consider, to be true.

Thus,

Thus, b_{m+1} is true if b_{m} is true

Hence, by mathematical induction,

For each natural no. n it is true that,

Question:19

A sequence Is defined by letting

Show that for all n ϵ N.

Answer:

A sequence Is defined by letting

Now, we'll substitute different values for k

Now, let us consider,

to be true.

Thus, is true if is true

Hence, by mathematical induction,

For each natural no. n it is true that,

Question:20

Answer:

Given data:

Now, we’ll substitute different values for n,

n = 1,

Thus, it’s true

Now, n = 2,

Thus, it’s true.

Now, let n = k

be true

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

Question:21

Answer:

Given:

Now, we’ll substitute different values for n,

Now, let us assume that,

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

Question:22

Answer:

Given:

P(n) :

Now, we’ll substitute different values for n=1,

Thus, it is true.Now,

let us assume that, for n=k is true

At n = k+1

Thus, n=k+1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

Question:23

Show that is a natural number for all n N.

Answer:

Given data:

P(n) =

Now, we’ll substitute different values for n,

At n=3

is true.

Let us consider,

Thus, at n=k+1

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

P(n) =

Question:24

Prove that ,for all natural numbers n > 1.

Answer:

Given:

P(n) =

Now, we’ll substitute different values for n,

.

It is true.

Now, let us consider,

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n > 1 it is true that,

P(n) =

Question:25

Prove that number of subsets of a set containing n distinct elements is , for all n N.

Answer:

To Prove:

is the no. of subsets of a set containing n distinct elements.

Now, there is only one subset since for a null set there is only one element φ

Now, we’ll substitute different values for n,

At n = 0,

No. of subsets = 1 =

If we consider a set that has 1 element then there will be 2 subsets that has 1 & φ

Thus, at n = 1, it is true

now at n=2

No. of subsets = 2 =

Now, let us consider that there is a set has k elements.

At n = k,

No. of subsets = is true.

Now, for set , having k+1 elements,

Compared to the set , has an extra element and thus will form an extra collection of subsets when combined with the existing subsets & an extra subset will be formed.

Thus, at n = k+1,

No. of subsets = +

= 2.

=

Thus, by mathematical Induction,

is the no. of subsets of a set containing n distinct elements.

Question:26

If is divisible by 9 for all n ? N, then the least positive integral value of k is

A. 5

B. 3

C. 7

D. 1

Answer:

The answer is the option (A) 5

Given:

P(n) = is divisible by 9

Now, we’ll substitute different values for n,

At n = 1,

Now, we know that sum of the digits of this no. should be 9 or multiples of 9, for it to be divisible by 9.

Thus

Thus,

Question:27

For all n ? N, is divisible by

A. 19

B. 17

C. 23

D. 25

Answer:

The answer is the option (B) 17

Given:

P(n) =

Now, we’ll substitute different values for n,

At n = 1,

P(1) =

At n = 2,

Therefore, it is divisible by 17.

Question:28

If is divisible by , then the least positive integral value of k is

A. 1

B. 2

C. 3

D. 4

Answer:

The answer is the option (A) 1

Given:

P(n) = is divisible by

Now, we’ll substitute different values for n,

At n = 1,

At n = 2,

At n = 3,

At n = 4,

Therefore, ‘1’ is the least possible integral value of k.

Question:29

Fill in the blanks in the following:

If , then P(n) is true for all __________.

Answer:

Given:

Now, we’ll substitute different values for n,

At n = 1,

,

, viz. not true.

At n = 2,

, viz. not true.

At n = 3,

, viz. not true

At n = 4,

, viz. not true

At n = 5,

, viz. true

Therefore,

Question:30

Answer:

True.

Explanation: It is the Principle of Mathematical Induction.

Deductive reasoning has helped in the field of mathematics to a great extent in understanding and deriving various conclusions out of already proven statements. It helps in establishing a new relationship out of two different things relating to a single entity. For example- if A is the father of B and also A is the son of G then we can establish a relationship between B and G that B is the grandson of G or G is the grandfather of B.

As per NCERT Exemplar Class 11 Maths solutions chapter 4, the same principle is used in mathematics for checking the relevance of certain statements through the process of deductive reasoning. It means applying any established principle of a general case to a special case to check its applicability and relevance.

This whole process is known as induction and is used in mathematics for proving different statements by using any established universal principle such as proving that a statement is true for a natural number x, then will be true for the next number x+1 as well as for verifying if any statement is true for a natural number y+1 then it will also apply to y.

Class 11 Maths NCERT Exemplar solutions chapter 4 are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams. The step-wise explanation of the chapter makes it easy for the students to understand the concept and method easily.

*This chapter has been omitted according to the revised syllabus of CBSE for 2020-21.

- Introduction
- Motivation
- The Principle of Mathematical Induction

What will the students learn from NCERT Exemplar Class 11 Maths solutions chapter 4?

NCERT Exemplar Class 11 Maths chapter 4 solutions is straightforward yet a difficult one but will ultimately help students to develop skills regarding inductive and deductive reasoning used in almost every other field ranging from science to many other branches.

NCERT Exemplar solutions for Class 11 Maths chapter 4 will help students to develop their skills relating to mathematical inductions that have its footsteps in every branch of science and is also a part of various competitive exams.

It also develops skills about establishing relationships between different things relating to some common term and is also used in daily routine for understanding relationships and connections between different things.

Important Topics to cover in NCERT Exemplar Class 11 Maths solutions chapter 4 Principle of Mathematical Induction

Although this chapter is omitted for the 2020-2021 examination, the process of inductive and deductive reasoning is important and so are the examples which you should pay attention to sharpen your mind and have more clarity on these types of questions.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | Principle of Mathematical Induction |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

**Also Check NCERT Books and NCERT Syllabus here:**

1. Who has prepared the solutions?

The NCERT Exemplar Class 11 Maths chapter 4 solutions are prepared by professional mathematicians on our team. The solutions are prepared after referring to various advanced-level books.

2. How many questions are there in NCERT Exemplar Class 11 Maths Solutions Chapter 4?

There are a total of 24 questions in one exercise. The questions are for problem-solving in this chapter.

3. How to prepare for maths exam?

You can practice maths by solving as many problems as you can. The more variety of questions you solve, the better prepared you are.

4. Are these solutions helpful for competitive examinations?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 4 are designed to develop your base and give you the best solutions to the competitive exam questions.

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