NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction in Hindi and English Medium updated for CBSE session 2022-2023.

## 11th Maths Chapter 4 Solutions in English Medium

### 11th Maths Chapter 4 Solutions in Hindi Medium

### Class 11 Maths Chapter 4 all Exercises Solution

### NCERT Solutions for Class 11 Maths Chapter 4

Class XI Ganit NCERT Chapter 4 Ganitiiy Aagman ka Siddhant Prashnavali 4.1 in Hindi Medium to view online in Video format Free download is available without login or password. All subjects in class 11 Offline Apps based on NCERT Solutions are available on Play Store and App Store. NCERT Solutions apps are based on latest textbooks for 11th grade given on NCERT (https://ncert.nic.in/) website following the new CBSE curriculum 2022-2023. Join the discussion forum to ask your doubts and share your knowledge with the others.

Class: 11 | Mathematics |

Chapter 4: | Principle of Mathematical Induction |

Content: | NCERT Exercises Solution |

Content Type: | Videos and Text format |

Medium: | English and Hindi Medium |

### 11th Maths Chapter 4 Solutions

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Exercise 4.1 in English & Hindi medium is given below for new academic year 2022-23. Visit to Class 11 Mathematics main page to get the solutions of all chapters.

These NCERT Solutions are useful for all the students CBSE Board, Uttarakhand Board, MP Board, UP Board, Bihar Board and all other board who are following NCERT Books based on new CBSE Curriculum 2022-23.

#### Class 11 Maths Chapter 4 Exercise 4.1 Solution in Videos

#### Class 11 Maths Exercise 4.1 Solution in Hindi

### Class 11 Maths Chatper 4 Question Answers

### What is meant by a statement in mathematics?

A meaningful sentence which can be judged to be either true or false is called a statement.

### What is a mathematical statement?

A statement involving mathematical relations is called as mathematical statement.

### What do know about Principle of mathematical Induction?

Principle of mathematical Induction:

1. Let P(n) be any statement involving natural number n such that

P(1) is true, and

2. If P(k) is true as well as P(k + 1) is true for some natural number k, i.e. P(k + 1) is true whenever P(k) is true for some natural number k, then P(n) is true for all natural numbers.

#### Important Terms on Mathematical Induction

Induction and deduction are two basic processes of reasoning.

1. Deduction: It is the application of a general case to a particular case. In contrast to deduction, induction is process of reasoning from particular to general.

2. Induction: Induction being with observations. From observations we arrive at tentative conclusions called conjectures. The process of induction help in proving the conjectures which may be true.

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### Important Questions on 11th Maths Chapter 4

### Prove the following by using the principle of mathematical induction for all n âˆˆ N: 1.2+ã€–2.2ã€—^2+ã€–3.2ã€—^3+â‹¯+n.2^n=(n-1) 2^(n+1)+2

Let the given statement be P(n), therefore,

P(n):1.2+ã€–2.2ã€—^2+ã€–3.2ã€—^3+â‹¯+n.2^n = (n-1) 2^(n+1)+2

For n = 1, we have L.H.S. = 1.2 = 2 R.H.S. = (1-1) 2^(1+1)+2 = 0+2=2

L.H.S. = R.H.S, so P(1) is true.

Let P(k) be true for some positive integer k, such that

P(k):1.2+ã€–2.2ã€—^2+ã€–3.2ã€—^3+â‹¯+k.2^k = (k-1) 2^(k+1)+2

Now, to prove that P(k + 1) is true. i.e. P(k+1):1.2+ã€–2.2ã€—^2+ã€–3.2ã€—^3+â‹¯+k.2^k+(k+1).2^(k+1) = (k) 2^(k+2)+2

Considering the L.H.S. of P(k + 1), we have 1.2+ã€–2.2ã€—^2+ã€–3.2ã€—^3+â‹¯+k.2^k+(k+1).2^(k+1)

= (1.2+ã€–2.2ã€—^2+ã€–3.2ã€—^3+â‹¯+k.2^k )+(k+1).2^(k+1)

= (k-1) 2^(k+1)+2+(k+1).2^(k+1) [As P(k) is true]

= (k-1) 2^(k+1)+(k+1).2^(k+1)+2

= [(k-1)+(k+1)].2^(k+1)+2 =2k.2^(k+1)+2

= k.2^(k+2)+2 =R.H.S.

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

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

### Prove the following by using the principle of mathematical induction for all n âˆˆ N: ã€–10ã€—^(2n-1)+1 is divisible by 11.

Let the given statement be P(n), therefore, P(n):ã€–10ã€—^(2n-1)+1 is divisible by 11.

For n = 1, we have ã€–10ã€—^(2-1)+1=11,which is divisible by 11.

So, P(1) is true.

Let P(k) be true for some positive integer k, such that P(k):ã€–10ã€—^(2k-1)+1 is divisible by 11.

Let ã€–10ã€—^(2n-1)+1 = 11m â€¦ (1) Where, m is any natural number.

Now, to prove that P(k + 1) is true. i.e. P(k+1):ã€–10ã€—^(2k+1)+1 is divisible by 11.

Consider ã€–10ã€—^(2k+1)+1 =ã€–10ã€—^(2k-1+2)+1

=ã€–10ã€—^2.ã€–10ã€—^(2k-1)+1

=ã€–10ã€—^2.(11m-1)+1 [From the equation (1),ã€–10ã€—^(2n-1) = 11m-1]

= 100.(11m-1)+1 =1100m â€“ 100 + 1

= 1100m â€“ 99

= 11[100m â€“ 9], which is divisible by 11.

Thus, P(k + 1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.

### Prove the following by using the principle of mathematical induction for all n âˆˆ N: x^2n-y^2n is divisible by x+y.

Let the given statement be P(n), therefore, P(n):xÂ²n-yÂ²n is divisible by x+y.

For n = 1, we have xÂ²-yÂ²=(x+y)(x-y), which is divisible by x+y.

So, P(1) is true.

Let P(k) be true for some positive integer k, such that P(k):xÂ²k-yÂ²k is divisible by x+y.

Let xÂ²k-yÂ²k=(x+y)m â€¦ (1) Where, m is any natural number.

Now, to prove that P(k + 1) is true. i.e. P(k+1):xÂ²(k+1) – yÂ²(k+1) is divisible by x+y.

Consider x^2(k+1) -y^2(k+1)

= x^(2k+2)-y^(2k+2) =ã€–x^2 xã€—^2k-y^(2k+2)

= xÂ² [(x+y)m+y^2k ]-y^(2k+2) [From the equation (1),x^2k=(x+y)m+y^2k ]

= xÂ² (x+y)m+x^2 y^2k-y^(2k+2)

= xÂ² (x+y)m+y^2k (xÂ²-yÂ²)

= xÂ² (x+y)m+y^2k (x-y)(x+y) =(x+y)[x^2 m+y^2k (x-y)], which is divisible by (x+y).

Thus, P(k + 1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.

### Prove the following by using the principle of mathematical induction for all n âˆˆ N: (2n+7)<(n+3)Â².

Let the given statement be P(n), therefore, P(n):(2n+7)<(n+3)Â².
For n = 1, we have (2Ã—1+7)<(1+3)Â²
â‡’ 9 < 16
â‡’ P(1) is true.
Let P(k) be true for some positive integer k, such that P(k):(2k+7)<(k+3)Â².
Now, to prove that P(k + 1) is true. i.e.
P(k+1):(2k+9)<(k+4)Â²
Considering the statement P(k), we have
(2k+7)<(k+3)Â²
[Difference between (k+4)Â² and (k+3)Â² = (k+4)Â² - (k+3)Â² = kÂ²+8k+16-(kÂ²+6k+9)=2k+7]
Adding both sides (2k+7), we have
(2k+7) + (2k+7)<(k+3)Â² + (2k+7)
â‡’(2k+9) + (2k+5) < (kÂ² +6k + 9) + (2k+7)
â‡’(2k+9)+(2k+5) < (kÂ² + 6k + 9 + 2k + 7)
â‡’(2k+9)+(2k+5) < (kÂ² +8k + 16)
â‡’(2k+9)+(2k+5) < (k+4)Â²
â‡’(2k+9)<(k+4)Â² [As k>0,so 2k+1>0]

Thus, P(k + 1) is true whenever P(k) is true. Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.

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### How many exercises are there in chapter 4 of grade 11th Maths?

There is only one exercise (Ex 4.1) having 24 questions and 8 illustrations in chapter 4 of class 11th Maths. This chapter is easy and unique. But, some examples and questions of this chapter are a little difficult.

### Is chapter 4 of class 11th Maths important for the school exams?

Yes, chapter 4 of class 11th Maths is important for the school exams. Every year 4 marks question is asked from chapter 4. Some important problems of this chapter are questions 1, 3, 5, 7, 8, 11, 13, 15, 16, 17, 18, 21, 23, 24, and examples 4, 5, 6, 7.

### Is there any need to practice extra questions of chapter 4 of class 11th Maths?

No, there is no need to practice extra questions of chapter 4 (Principle of Mathematical Induction) of class 11th Maths. NCERT questions, Exemplar book and examples are sufficient. But, if someone wants to practice extra questions of chapter 4, they can practice extra questions from R.L. Arora, R.S. Aggarwal, R.D. Sharma, etc.

### Can students complete chapter 4 of class 11th Maths in a day?

No, students cannot complete chapter 4 of class 11th Maths in 1 day. Students need a maximum of 3-4 days if they seriously give 1-2 hours per day to this chapter. This time also depends on studentâ€™s efficiency, capability, working speed, etc.