Mathematical Induction Ncert Solutions Pdf

*According to the Revised CBSE Syllabus 2020-21, this chapter has been removed.

NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction are given in an understandable way by the faculty at CoolGyan'S. Students learn about the Principle of Mathematical Induction and its application in detail through this chapter. By practising all the problems present in the NCERT Solutions, students can easily score maximum marks in the examinations.
Principle of Mathematical Induction is a specific technique used to prove certain mathematically accepted statements in algebra and in other applications of Mathematics, such as inductive and deductive reasoning. NCERT Solutions of CoolGyan'S cover all these concepts and help in scoring full marks in this chapter. These solutions are useful for further studies and for those who are preparing for competitive exams. NCERT Solutions For Class 11 Maths are very accurate and make it easy for the students to crack the exam with good marks.

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Exercise 4.1 Solutions 24 Questions

Access NCERT Solutions for Class 11 Maths Chapter 4

Exercise 4.1 page: 94

Prove the following by using the principle of mathematical induction for all n ∈ N:

1.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

2.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

3.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

4.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

5.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

6.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

7.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

8. 1.2 + 2.2² + 3.2² + … +n.2 ⁿ  = (n – 1) 2 ^{n +1} + 2

Solution:

We can write the given statement as

P (n): 1.2 + 2.2² + 3.2² + … +n.2 ⁿ  = (n – 1) 2 ^{n +1} + 2

If n = 1 we get

P (1): 1.2 = 2 = (1 – 1) 2¹⁺¹ + 2 = 0 + 2 = 2

Which is true.

Consider P (k) be true for some positive integer k

1.2 + 2.2² + 3.2² + … +k.2 ^k  = (k – 1) 2 ^{k  + 1} + 2 … (i)

Now let us prove that P (k + 1) is true.

Here

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

9.

Solution:

We can write the given statement as

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

10.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

11.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

12.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

13.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

14.

Solution:

By further simplification

= (k + 1) + 1

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

15.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

16.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

17.

Solution:

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

18.

Solution:

We can write the given statement as

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

19. n (n + 1) (n + 5) is a multiple of 3

Solution:

We can write the given statement as

P (n): n (n + 1) (n + 5), which is a multiple of 3

If n = 1 we get

1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

Which is true.

Consider P (k) be true for some positive integer k

k (k + 1) (k + 5) is a multiple of 3

k (k + 1) (k + 5) = 3m, where m ∈ N …… (1)

Now let us prove that P (k + 1) is true.

Here

(k + 1) {(k + 1) + 1} {(k + 1) + 5}

We can write it as

= (k + 1) (k + 2) {(k + 5) + 1}

By multiplying the terms

= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)

So we get

= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)

Substituting equation (1)

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

By multiplication

= 3m + (k + 1) {2k + 10 + k + 2}

On further calculation

= 3m + (k + 1) (3k + 12)

Taking 3 as common

= 3m + 3 (k + 1) (k + 4)

We get

= 3 {m + (k + 1) (k + 4)}

= 3 × q where q = {m + (k + 1) (k + 4)} is some natural number

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

20. 10^{2 n  – 1}+ 1 is divisible by 11

Solution:

We can write the given statement as

P (n): 10^{2 n  – 1}+ 1 is divisible by 11

If n = 1 we get

P (1) = 10^{2.1 – 1}+ 1 = 11, which is divisible by 11

Which is true.

Consider P (k) be true for some positive integer k

10^{2 k  – 1}+ 1 is divisible by 11

10^{2 k  – 1}+ 1 = 11m, wherem ∈ N…… (1)

Now let us prove that P (k + 1) is true.

Here

10 ^{2 (k + 1) – 1} + 1

We can write it as

= 10 ^{2k + 2 – 1} + 1

= 10 ^{2k + 1} + 1

By addition and subtraction of 1

= 10 ² (10^2k-1 + 1 – 1) + 1

We get

= 10 ² (10^2k-1 + 1) – 10² + 1

Using equation 1 we get

= 10². 11m – 100 + 1

= 100 × 11m – 99

Taking out the common terms

= 11 (100m – 9)

= 11 r, where r = (100m – 9) is some natural number

10 ^{2(k + 1) – 1} + 1 is divisible by 11

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

21. x ^{2 n}  –y ^{2 n}  is divisible by x+y

Solution:

We can write the given statement as

P (n):x ^{2 n}  –y ^{2 n}  is divisible by x+y

If n = 1 we get

P (1) = x ^2 × 1 –y ^2 × 1 =x ² –y ² = (x+y) (x –y), which is divisible by (x + y)

Which is true.

Consider P (k) be true for some positive integer k

x ^{2 k}  –y ^{2 k}  is divisible by x+y

x ^{2 k}  –y ^{2 k}  =m (x+y), wherem ∈ N…… (1)

Now let us prove that P (k + 1) is true.

Here

x ^{2(k + 1)} – y ^{2(k + 1)}

We can write it as

= x ^2k . x² – y^2k . y²

By adding and subtracting y^2k we get

= x² (x^2k – y^2k + y^2k) – y^2k. y²

From equation (1) we get

= x² {m (x + y) + y^2k} – y^2k. y²

By multiplying the terms

= m (x + y) x² + y^2k. x² – y^2k. y²

Taking out the common terms

= m (x + y) x² + y^2k (x² – y²)

Expanding using formula

= m (x + y) x² + y^2k (x + y) (x – y)

So we get

= (x + y) {mx² + y^2k (x – y)}, which is a factor of (x + y)

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

22. 3^{2 n  + 2} – 8n – 9 is divisible by 8

Solution:

We can write the given statement as

P (n): 3^{2 n  + 2} – 8n – 9 is divisible by 8

If n = 1 we get

P (1) = 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8

Which is true.

Consider P (k) be true for some positive integer k

3^{2 k  + 2} – 8k – 9 is divisible by 8

3^{2 k  + 2} – 8k – 9 = 8m, wherem ∈ N…… (1)

Now let us prove that P (k + 1) is true.

Here

3 ^{2(k + 1) + 2} – 8 (k + 1) – 9

We can write it as

= 3 ^{2k + 2} . 3² – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 3² (3^{2k + 2} – 8k – 9 + 8k + 9) – 8k – 17

On further simplification

= 3² (3^{2k + 2} – 8k – 9) + 3² (8k + 9) – 8k – 17

From equation (1) we get

= 9. 8m + 9 (8k + 9) – 8k – 17

By multiplying the terms

= 9. 8m + 72k + 81 – 8k – 17

So we get

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

So 3 ^{2(k + 1) + 2} – 8 (k + 1) – 9 is divisible by 8

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

23. 41 ⁿ  – 14 ⁿ  is a multiple of 27

Solution:

We can write the given statement as

P (n):41 ⁿ  – 14 ⁿ is a multiple of 27

If n = 1 we get

P (1) = 41¹ – 14¹ = 27, which is a multiple by 27

Which is true.

Consider P (k) be true for some positive integer k

41 ^k  – 14 ^k is a multiple of 27

41 ^k  – 14 ^k  = 27m, wherem ∈ N…… (1)

Now let us prove that P (k + 1) is true.

Here

41^{k + 1} – 14 ^{k + 1}

We can write it as

= 41^k. 41 – 14^k. 14

By adding and subtracting 14^k we get

= 41 (41^k – 14^k + 14^k) – 14^k. 14

On further simplification

= 41 (41^k – 14^k) + 41. 14^k – 14^k. 14

From equation (1) we get

= 41. 27m + 14^k ( 41 – 14)

By multiplying the terms

= 41. 27m + 27. 14^k

By taking out the common terms

= 27 (41m – 14^k)

= 27r, where r = (41m – 14^k) is a natural number

So 41^{k + 1} – 14^{k + 1} is a multiple of 27

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

24. (2n+7) < (n + 3)²

Solution:

We can write the given statement as

P(n): (2n+7) < (n + 3)2

If n = 1 we get

2.1 + 7 = 9 < (1 + 3)² = 16

Which is true.

Consider P (k) be true for some positive integer k

(2k + 7) < (k + 3)² … (1)

Now let us prove that P (k + 1) is true.

Here

{2 (k + 1) + 7} = (2k + 7) + 2

We can write it as

= {2 (k + 1) + 7}

From equation (1) we get

(2k + 7) + 2 < (k + 3)² + 2

By expanding the terms

2 (k + 1) + 7 < k² + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k² + 6k + 11

Here k² + 6k + 11 < k² + 8k + 16

We can write it as

2 (k + 1) + 7 < (k + 4)²

2 (k + 1) + 7 < {(k + 1) + 3}²

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

---

NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction

This chapter has only one exercise which will help students in understanding the concepts related to the Principle of Mathematical Induction clearly. The major topic and subtopics covered in Chapter 4  Principle of Mathematical Induction of NCERT Solutions for Class 11 include the following.
4.1 Introduction
Here, students can understand deductive reasoning with suitable examples. This section explains the assumptions that are made on the basis of certain universal facts.
4.2 Motivation
In this section, mathematical induction is explained with a real-life scenario to make the students understand how it basically works.
4.3 The Principle of Mathematical Induction
This section explains the Principle of Mathematical Induction using inductive step and the inductive hypothesis.
Suppose there is a given statement P(n) involving the natural number n such that

• The statement is true for n = 1, i.e., P(1) is true
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., the truth of P(k) implies the truth of P (k + 1).

Key Features of NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction

Studying the Principle of Mathematical Induction of Class 11 enables the students to understand the process of the proof by induction, motivating the application of the method by looking at natural numbers as the least inductive subset of real numbers. Students also get to know the principle of mathematical induction and its applications after going through the solutions of NCERT questions. The summary of the concepts discussed and used in the solutions of this chapter are:

1. One key basis for mathematical thinking is deductive reasoning. In contrast to deduction, inductive reasoning depends on working with different cases and developing a conjecture by observing incidences until we have observed each and every case. Thus, in simple language we can say the word 'induction' means the generalization from particular cases or facts
2. The principle of mathematical induction is one such tool which can be used to prove a wide variety of mathematical statements. Each such statement is assumed as P(n) associated with positive integer n, for which the correctness of the case n = 1 is examined. Then assuming the truth of P(k) for some positive integer k, the truth of P (k+1) is established.

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 4

What is the concept of Mathematical Induction explained in Chapter 4 of NCERT Solutions for Class 11 Maths?

Mathematical Induction is basically the method of confirming statements in order to prove a statement, formula or theorem which is true for all the natural numbers. There are two steps which are involved in proving a statement.
1. Base step.
2. Inductive step.

Explain the first principle of Mathematical Induction in Chapter 4 of NCERT Solutions for Class 11 Maths.

The first principle of Mathematical Induction means that when both the base step and inductive step are true, we can conclude that the initial statement is true for all natural numbers. Students can find numerous problems based on this concept from the NCERT textbook of Class 11. Solutions for the exercise wise problems are also available in PDF format with the aim of helping students to score well in the annual exam.

Why is the NCERT Solutions for Class 11 Maths Chapter 4 PDF important from the exam point of view?

The NCERT Solutions for Class 11 Maths Chapter 4 provides a clear idea about the induction and deduction techniques which are used to prove equations and statements. By using the solutions PDF, students will get an in-depth knowledge about the principle of Mathematical Induction and its applications. These solutions will improve confidence among students to prepare for the Class 11 exams.

Mathematical Induction Ncert Solutions Pdf

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