NCERT Solutions Class 11 Maths Chapter 4

NCERT Solutions Class 11 Mathematics Chapter 4 – Principle of Mathematical Induction

NCERT solution Class 11 Mathematical  Chapter 4- Principle of Mathematical Induction is available on the Extramarks website for exam preparation. The consistent practice of the solutions will offer an in-depth understanding of the base. Furthermore, the examples in these solutions are perfect for learning core concepts related to the proof by induction and its applications. Thus, students can refer to the NCERT Solutions Class 11 Mathematics Chapter 4. 

The principle of Mathematical  Induction and reasoning depends on working with each case and developing an assumption by observing incidences until we have the application’s deduction. In NCERT solution Class 11 Mathematics Chapter 4 helps the students to find out how to use Mathematical induction to prove many statements. Moreover, they will understand how to calculate the inductive reasoning.

Students can get the NCERT solution Class 11 Mathematics Chapter 4 and other NCERT Solutions on our Extramarks website. 

 

Key Topics Covered In NCERT Solution for Class 11 Mathematics Chapter 4

The Class 11 NCERT Mathematics Solutions Chapter 4 introduces the deductive process, motivators and the concept of mathematical reasoning. The formulas, along with relevant examples, are included in each subject to provide an understanding of the concepts.

The NCERT Solution for Class 11 Mathematics Chapter 4 consists of 24 questions, of which the difficulty level ranges from easy to complex. A few of these questions require a thorough analysis and thinking method to develop solutions. There are two fundamental methods to solve questions based on mathematical induction. First, to determine whether the statement is true in the specific case, we need the formulas of a particular series, which are given below.

Then, in these NCERT Solutions Class 11 Mathematics Chapter 4 Principles of Mathematical Induction, various properties and mathematical concepts that form the foundation of the maths theory are explained in depth.

Exercise  Topics
4.1 

4.2

4.3

 Introduction

Motivation

The Principle of Mathematical Induction

 

There are a total of three parts to this exercise. First, a summary of the topics and sub-topics is covered in NCERT Solutions Class 11 Mathematics Chapter 4:

 

Part 1- Introduction

The students will get an introduction to the Mathematical Induction. Mathematical induction is a term which helps prove mathematical proofs and theorems that apply to any natural number. The concept of mathematical induction is a particular method used to prove certain assertions in algebra, which are constructed by n, in which n is an unnatural number. For example, every mathematical expression, statement or equation is proven based on the assumption that it is valid for n = 1.  The n value is k, and the proof is then made by comparing the case of n = 1 + k.

The initial portion of the chapter teaches a fundamental understanding that deductive reasoning is a method of thinking. Questions in this section comprise at least two sentences that outline an argument or logic question that students have to prove true or false with the help of deductive steps.

 

Part 2- Motivation

In this portion, students are introduced to a key idea that they should know before solving the mathematical sums induction. The principle of motivation involves investigating the proof that the statement you are proving is true for a particular natural number; it’s also true for all natural numbers. 

Let’s suppose there’s an assertion P(n) based on an unnatural number in a sense that.

The statement is true when the case of n = 1, i.e., P(1) is valid.

If the statement is true, n = k, which is a naturally occurring number, this statement is also valid when n is k+1. i.e. that the reality of P(k) indicates what is true of P(k+1)

In this case, P(n) is valid for all natural numbers with n.

This is the basis for the principle that is the foundation of Mathematical Induction. It is, therefore, crucial to get a solid grasp on this subject so that you can solve the related problems.

 

Illustration

This is a section where many students struggle to use the correct logic to prove the validity of the  given statements. This is the  reason why NCERT Solution Class 11 Mathematics Chapter 4 can be useful for students to discover the best method of dealing with equations while deducting them. .

 

Part 3- The Principle of Mathematical Induction

This section will guide the students through the concept of Mathematical Induction with the help of the concepts mentioned earlier. This means that they will have to work out equations using applications to the exact.

Each step utilised in proving the theory assertion using mathematical induction is given an established name. Every step is identified in the following manner:

  • The Base step: To demonstrate that P(1) is correct.
  • Assumption step: Assuming that P(k) is valid for a certain K in N.
  • Introduction step: Prove that P(k+1) is the case.

 

NCERT Solutions Class 11 Mathematics Chapter 4: Exercise & Solutions

Induction in mathematics can be used to draw the results and proofs that help to grasp the mathematical concepts and theorems. The answers in these NCERT Solutions for Class 11 Mathematics Chapter 4: Principles of Mathematical Induction, various properties and mathematical concepts which are the basis of the mathematical theory are explained in depth. The students can take advantage of the NCERT solution Class 11 Mathematics Chapter 4. 

NCERT Solutions Class 11 Mathematics Chapter 4 covers a brief introduction to deductive reasoning, motivation and the principle of mathematical induction. It also has relevant formulas with appropriate examples provided on each topic. 

By utilising this NCERT Solutions Class 11 Mathematics Chapter 4 Principle of Mathematical Induction, students can effectively develop a simplified method for solving problems. The well-designed layout of these solutions efficiently enhances the basic concepts needed for successful exam preparation—a thorough analysis of exercise-based solutions.

 

The students can click on the  links given below to refer to exercise questions and solutions for NCERT solution Class 11 Mathematics Chapter 4: 

  • Class 11 Mathematics Chapter 4 Ex 4.1

 

Along with this, students can also refer to the NCERT solutions Class Mathematics Chapter 4. 

 

NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions for Class 9, Class 10, and Class 11

 

NCERT Exemplar Class 11 Mathematics 

NCERT Exemplar is the most effective study material to test knowledge about the subject covered in the chapter. It contains short-form questions, lengthy-form questions, as well as multiple-choice questions. Additionally, the Exemplar includes numerous examples with detailed solutions. The Exemplar’s practice can help students strengthen their understanding of Chapter 4 Mathematics Class 11. 

Additionally, they can increase their understanding and abilities regarding the subject. The difficulty level varies from the subject covered in the chapter. For instance, the difficulty level of the Principle of Mathematical Induction Chapter 4 is moderate. Therefore, it’s appropriate for Class 11 and above.  

This chapter focuses on the basic idea of deductive reasoning. The questions in the section consist of two or more sentences. Further, the students need to solve problems by stating an argument or a logical question that students need to prove as true or false. The deductive steps and guidelines will help to solve the questions. This chapter is important because it will allow the students to assess their ability to apply their knowledge to induction. For Class 11 students,  the  NCERT  Exemplar provides a solid  Mathematical foundation along with other subjects.

 For better performance, students can get NCERT Solutions Class 11 Mathematics Chapter 4 and other primary NCERT solutions. It includes NCERT solutions Class 1, NCERT solutions Class 2, NCERT solutions Class 3, NCERT solutions Class 4 and NCERT solutions Class 5. 

 

Key Features of NCERT Solutions Class 11 Mathematics Chapter 4

 To improve your score, it is essential to practise the important chapters and key topics. Thus, students can refer to NCERT Solutions Class 11 Mathematics Chapter 4. Some of the key features include:

  • The solutions provide an in-depth guideline on the methods of deduction and induction employed to demonstrate equations and statements.
  • The answers also include possible questions and answers developed through extensive research by our experts in the field.
  • It can help students understand fundamental concepts through clever problem-solving methods.
  • The detailed answers will assist the students in studying and practising Mathematics.

Q.1 Prove by using principal of mathematical induction for all n ∈ N

1+3+32+...+3n–1=(3n−1)2.

Ans

Let P(n) be the given statement,i.e.,  P(n):1+3+32+...+3n–1=(3n−1)2For n =1,P(1)  =1R.H.S.=31−12=1=L.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1+3+32+...+3k–1=(3k−1)2 ...(1)We need to prove that P(k + 1) is true whenever P(k) is true.We have             1+3+32+...+3k–1+3k=(3k+1−1)2⇒     (3k−1)2+3k=(3k+1−1)2 ...(2) [From equation (1)]L.H.S.:     (3k−1)2+3k=(3k−1+2.3k)2       =3.3k−12       =3k+1−12=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.2 Prove by using principal of mathematical induction for all n ∈ N

13+23+33+...+n3={n(n+1)2}2.

Ans

Let P(n) be the given statement,i.e.,  P(n):13+23+33+...+n3={n(n+1)2}2For n =1,P(1)=13=1R.H.S.:{1.(1+1)2}2=1.∴L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):13+23+33+...++nk={k(k+1)2}2...(i)We need to prove that P(k+1) is true whenever P(k) is true.We have  13+23+33+...++k3+(k+1)3={(k+1)(k+2)2}2⇒    {k(k+1)2}2+(k+1)3={(k+1)(k+2)2}2 ...(2)[From equation(i)]L.H.S.:    {k(k+1)2}2+(k+1)3=(k+1)2{k4+k+1}                 =(k+1)2{k2+4k+44}                =(k+1)2{(k+2)24}                ={(k+1)(k+2)2}2=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.3 Prove by using principal of mathematical induction for all n ∈ N

1+1(1+2)+1(1+2+3)+...+1(1+2+3+4+...+n)=2nn+1.

Ans

Let P(n) be the given statement,i.e.,  P(n):1+1(1+2)+1(1+2+3)+...+1(1+2+3+4+...+n)=2nn+1For n =1,  ⇒P(1)=1R.H.S.:2(1)1+1=1.∴           L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1+1(1+2)+1(1+2+3)+...+1(1+2+3+4+...+k)     =2kk+1 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have1+1(1+2)+1(1+2+3)+...+1(1+2+3+4+...+k)+1{1+2+3+4+...+(k+1)}=2(k+1)k+2⇒               2kk+1+1{1+2+3+4+...+(k+1)}=2(k+1)k+2 ...(2)[From equation(1)]L.H.S. : 2kk+1+1{1+2+3+4+...+(k+1)} =2kk+1+1{(k+1)(k+1+1)2}        =2kk+1+2(k+1)(k+2)        =2k+1(k+2k+2)        =2k+1(k2+2k+1k+2)        =2k+1×(k+1)2k+2        =2(k+1)(k+1)+1=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.4 Prove by using principal of mathematical induction for all n ∈ N

1.2.3+2.3.4+...+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4.

Ans

Let P(n) be the given statement,i.e.,  P(n):1.2.3+2.3.4+...+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4For n =1,P(1)=1.2.3=6R.H.S.:       1(1+1)(1+2)(1+3)4=1(2)(3)(4)4         =6∴L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1.2.3+2.3.4+...+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have1.2.3+2.3.4+...+k(k+1)(k+2) +(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)4⇒k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)4 ...(2)[From equation (1)]L.H.S.:  k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k4+1) =(k+1)(k+2)(k+3)(k+4)4=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, byprinciple of mathematical induction, P(n) is true for every positive integer n.

Q.5 Prove by using principal of mathematical induction for all n ∈ N

1.3+2.32+3.33+...n.3n=(2n–1)3n+1+34

Ans

Let P(n) be the given statement,i.e.,  P(n):1.3+2.32+3.33+...+n.3n=(2n−1)3n+1+34For n =1,P(1)=1.3=3R.H.S.:  (2×1−1)31+1+34=124       =3∴L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1.3+2.32+3.33+...+k.3k=(2k−1)3k+1+34 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have1.3+2.32+3.33+...+k.3k +(k+1)3k+1={2(k+1)−1}3k+2+34⇒    (2k−1)3k+1+34+(k+1)3k+1=(2k+1)3k+2+34  …(2)[From equation(1)]L.H.S.:  (2k−1)3k+1+34+(k+1)3k+1=(2k−1)3k+1+34+(k+1)3k+1 =(2k−1)3k+1+3+4(k+1)3k+14 ={2k−1+4(k+1)}3k+1+34 ={2k−1+4k+4}3k+1+34 ={6k+3}3k+1+34 =(2k+1)3k+2+34=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.6 Prove by using principal of mathematical induction for all n ∈ N

1.2+2.3+3.4+…+n.(n+1)=n(n+1)(n+2)3.

Ans

Let P(n) be the given statement,i.e.,  P(n): 1.2+2.3+3.4+…+n.(n+1)=n(n+1)(n+2)3For n =1,P(1)=1.2=2R.H.S.:1(1+1)(1+2)3=1(2)(3)3 =2∴L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1.2+2.3+3.4+…+k.(k+1)=k(k+1)(k+2)3 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have1.2+2.3+3.4+…+k.(k+1) +(k+1)(k+2)         =(k+1)(k+2)(k+3)3⇒     k(k+1)(k+2)3+(k+1)(k+2)=(k+1)(k+2)(k+3)3 ...(2)           [From equation(1)]L.H.S.:  k(k+1)(k+2)3+(k+1)(k+2)=(k+1)(k+2)(k3+1)       =(k+1)(k+2)(k+3)3       =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.7 Prove by using principal of mathematical induction for all n ∈ N

1.3+3.5+5.7+...+(2n−1)(2n+1)=n(4n2+6n−1)3.

Ans

Let P(n) be the given statement,i.e.,  P(n): 1.3+3.5+5.7+...+(2n−1)(2n+1)=n(4n2+6n−1)3For n =1,P(1)=1.3=3R.H.S. :   1.(4×12+6×1−1)3=1(9)3                             =3∴     L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1.3+3.5+5.7+...+(2k−1)(2k+1)                             =k(4k2+6k−1)3 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have1.3+3.5+5.7+...+(2k−1)(2k+1)+ (2k+1)(2k+3)                             =(k+1){4(k+1)2+6(k+1)−1}3⇒k(4k2+6k−1)3+(2k+1)(2k+3)=(k+1){4(k2+2k+1)+6k+6−1}3           [From equation(1)]⇒k(4k2+6k−1)3+(2k+1)(2k+3)=(k+1)(4k2+8k+4+6k+6−1)3⇒k(4k2+6k−1)3+(2k+1)(2k+3)=(k+1)(4k2+14k+9)3 ...(2)L.H.S.:  k(4k2+6k−1)3+(2k+1)(2k+3)=k(4k2+6k−1)3+4k2+6k+2k+3=4k3+6k2−k+12k2+18k+6k+93=4k3+18k2+23k+93=(k+1)(4k2+14k+9)3=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.8 Prove by using principal of mathematical induction for all n ∈ N

1.2+2.22+3.22+...+n.2n=(n–1)2n+1+2.

Ans

Let P(n) be the given statement,i.e.,  P(n): 1.2+2.22+3.22+...+n.2n=(n−1)2n+1+2For n =1,P(1):L.H.S.=1.2=2R.H.S.:(1−1)21+1+2=2∴       L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):1.2+2.22+3.22+...+k.2k=(k−1)2k+1+2 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have1.2+2.22+3.22+...+k.2k+ (k+1).2k+1 =(k+1−1)2k+1+1+2⇒       (k−1)2k+1+2+(k+1).2k+1=k.2k+2+2 ...(2)   [From equation(1)]L.H.S.:           (k−1)2k+1+2+(k+1).2k+1=(k−1+k+1)2k+1+2            =(2k)2k+1+2            =k.2k+2+2            =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.9 Prove by using principal of mathematical induction for all n ∈ N

12+14+18+...+12n=1–12n.

Ans

Let P(n) be the given statement,i.e.,  P(n): 12+14+18+...+12n=1−12nFor n =1,P(1)=12R.H.S.:1−121=12∴     L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):12+14+18+...+12k=1−12k ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have12+14+18+...+12k+ 12k+1 =1−12k+1⇒               1−12k+12k+1=1−12k+1 ...(2)     [From equation(1)]L.H.S.:               1−12k+12k+1=1−(2−12k+1) =1−12k+1 =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.10 Prove by using principal of mathematical induction for all n ∈ N

12.5+15.8+18.11+...+1(3n–1)(3n+2)=n(6n+4).

Ans

Let P(n) be the given statement,i.e.,  P(n): 12.5+15.8+18.11+...+1(3n–1)(3n+2)=n(6n+4)For n =1,P(1):       L.H.S.=12.5=110R.H.S.: 1(6×1+4)=110∴      L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):12.5+15.8+18.11+...+1(3k−1)(3k+2)=k(6k+4) ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have12.5+15.8+18.11+...+1(3k−1)(3k+2)+ 1{3(k+1)−1}{3(k+1)+2}                      =k+1{6(k+1)+4}⇒                         k(6k+4)+1(3k+2)(3k+5)=k+1(6k+10) ...(2)   [From equation (1)]L.H.S.:                                 k(6k+4)+1(3k+2)(3k+5)=3k2+5k+22(3k+2)(3k+5) =(3k+2)(k+1)2(3k+2)(3k+5) =k+1(6k+10)=R.H.S. Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for everypositive integer n.

Q.11 Prove by using principal of mathematical induction for all n ∈ N

11.2.3+12.3.4+13.4.5+...+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)

Ans

Let P(n) be the given statement,i.e.,  P(n): 11.2.3+12.3.4+13.4.5+...+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)For n =1,P(1):   11.2.3=16R.H.S.:1(1+3)4(1+1)(1+2)=44×2×3=16∴       L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k, i.e., We haveP(k):11.2.3+12.3.4+13.4.5+...+1k(k+1)(k+2)     =k(k+3)4(k+1)(k+2)   ...(1) We need to prove that P(k+1) is true whenever P(k) is true.11.2.3+12.3.4+13.4.5+...+1k(k+1)(k+2)+1(k+1)(k+2)(k+3)     =(k+1)(k+4)4(k+2)(k+3)⇒k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3)     =(k+1)(k+4)4(k+2)(k+3)   ...(2)   [From equation (1)]L.H.S.: k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3)     =k(k+3)2+44(k+1)(k+2)(k+3)     =k(k2+6k+9)+44(k+1)(k+2)(k+3)     =k3+6k2+9k+44(k+1)(k+2)(k+3)     =(k+1)(k2+5k+4)4(k+1)(k+2)(k+3)     =(k+1)(k+1)(k+4)4(k+1)(k+2)(k+3)     =(k+1)(k+4)4(k+2)(k+3)     =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.12 Prove by using principal of mathematical induction for all n ∈ N

a+ar+ar2+ar3+...+arn–1=a(rn−1)r−1

Ans

Let P(n) be the given statement,i.e.,     P(n): a+ar+ar2+ar3+...+arn–1=a(rn−1)r−1For n =1,P(1):     L.H.S.=  aR.H.S.:       a(r1−1)r−1=a∴          L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):   a+ar+ar2+ar3+...+ark–1=a(rk−1)r−1 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have     a+ar+ar2+ar3+...+ark–1+ark=a(rk+1−1)r−1⇒       a(rk−1)r−1+ark=a(rk+1−1)r−1  ...(2)    [From equation(1)]L.H.S.:                a(rk−1)r−1+ark=a(rk−1)+ark(r−1)r−1        =ark−a+ark+1−arkr−1        =a(rk+1−1)r−1        =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.13 Prove by using principal of mathematical induction for all n ∈ N

(1+31)(1+54)(1+79)...(1+(2n+1)n2)=(n+1)2.

Ans

Let P(n) be the given statement,i.e.,     P(n):(1+31)(1+54)(1+79)...{1+(2n+1)n2}=(n+1)2For n =1,P(1)=  (1+31)=4R.H.S.:(1+1)2=4∴         L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e.,P(k):(1+31)(1+54)(1+79)...{1+(2k+1)k2}=(k+1)2  …(1)We need to prove that P(k+1) is true whenever P(k) is true.We have(1+31)(1+54)(1+79)...{1+(2k+1)k2}{1+(2k+3)(k+1)2}=(k+2)2⇒                      (k+1)2{1+(2k+3)(k+1)2}=(k+2)2  …(2)    [From equation(1)]L.H.S.: (k+1)2{1+(2k+3)(k+1)2}=(k+1)2{(k+1)2+(2k+3)(k+1)2}     =k2+2k+1+2k+3     =k2+4k+4     =(k+2)2     =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.14 Prove by using principal of mathematical induction for all n ∈ N

(1+11)(1+12)(1+13)...(1+1n)=(n+1).

Ans

Let P(n) be the given statement,i.e.,     P(n):(1+11)(1+12)(1+13)...(1+1n)=(n+1)For n =1,P(1)=  (1+11)=2R.H.S.:(1+1)=2∴      L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):(1+11)(1+12)(1+13)...(1+1k)=(k+1) ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have(1+11)(1+12)(1+13)...(1+1k)(1+1k+1)=(k+2)⇒(k+1)(1+1k+1)=(k+2) ...(2)   [From equation(i)]L.H.S.:(k+1)(1+1k+1)=(k+1)(k+2k+1)        =(k+2)        =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.15 Prove by using principal of mathematical induction for all n ∈ N

12+32+52+...+(2n–1)2=n(2n–1)(2n+1)3

Ans

Let P(n) be the given statement,i.e.,     P(n):12+32+52+...+(2n−1)2=n(2n−1)(2n+1)3For n =1,P(1)=  12=1R.H.S. :1(2×1−1)(2×1+1)3=1∴      L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):12+32+52+...+(2k−1)2=k(2k−1)(2k+1)3 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.We have12+32+52+...+(2k−1)2+(2k+1)2=(k+1)(2k+1)(2k+3)3⇒         k(2k−1)(2k+1)3+(2k+1)2=(k+1)(2k+1)(2k+3)3 ...(2)   [From equation(1)]L.H.S.:  k(2k−1)(2k+1)3+(2k+1)2=(2k+1){k(2k−1)3+(2k+1)}   =(2k+1){2k2−k+6k+33}   =(2k+1){2k2+5k+33}   =(2k+1){(2k+3)(k+1)3}   =(k+1)(2k+1)(2k+3)3   =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, byprinciple of mathematical induction, P(n) is true for every positive integer n.

Q.16 Prove by using principal of mathematical induction for all n ∈ N

11.4+14.7+17.10+...+1(3n−2)(3n+1)=n(3n+1)

Ans

Let P(n) be the given statement, i.e.,     P(n):11.4+14.7+17.10+...+1(3n−2)(3n+1)=n(3n+1)For n =1,P(1)=  11.4=14R.H.S.:1(3×1+1)=14∴      L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):   11.4+14.7+17.10+...+1(3k−2)(3k+1)=k(3k+1)  …(1)We need to prove that P(k+1) is true whenever P(k) is true.We have11.4+14.7+17.10+...+1(3k−2)(3k+1)+1(3k+1)(3k+4)=k+1(3k+4)⇒  k(3k+1)+1(3k+1)(3k+4)=k+1(3k+4) ...(2)[From equation(1)]L.H.S. : k(3k+1)+1(3k+1)(3k+4)=1(3k+1)(k+13k+4)               =1(3k+1)(3k2+4k+13k+4)               =1(3k+1){(3k+1)(k+1)3k+4}               =k+1(3k+4)=R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.17 Prove by using principal of mathematical induction for all n ∈ N

13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)

Ans

Let P(n) be the given statement,i.e.,     P(n):13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)For n =1,   P(1):        L.H.S.=  13.5=115R.H.S.: 13(2×1+3)  =115∴         L.H.S.=R.H.S.Thus, P(n) is true for n=1.Let P(k) is true for some natural number k,i.e., P(k):   13.5+15.7+17.9+...+1(2k+1)(2k+3)=k3(2k+3)  …(1)We need to prove that P(k+1) is true whenever P(k) is true.We have13.5+15.7+17.9+...+1(2k+1)(2k+3)+1(2k+3)(2k+5)=k+13(2k+5)⇒ k3(2k+3)+1(2k+3)(2k+5)=k+13(2k+5)   …(2)  [From equation (i)]L.H.S. : k3(2k+3)+1(2k+3)(2k+5)=1(2k+3){k3+12k+5}                     =1(2k+3){2k2+5k+33(2k+5)}                     =1(2k+3){(k+1)(2k+3)3(2k+5)}                     =(k+1)3(2k+5)                     =R.H.S.Therefore, P(k + 1) is true when P(k) is true. Hence, byprinciple of mathematical induction, P(n) is true for everypositive integer n.

Q.18 Prove by using principal of mathematical induction for all n ∈ N

1+2+3+...+n<18(2n+1)2

Ans

Let P(n) be the given statement, i.e.,P(n):1+2+3+...+n<18(2n+1)2P(n) is true when n=1, since 1<18(2×1+1)2⇒1<98Thus, P(n) is true for n=1.Let P(n) is true for some natural number k, i.e.,P(k): 1+2+3+...+k<18(2k+1)2...(1)We need to prove that P(k+1) is true whenever P(k) is true.Adding (k+1) both sides of relation(1), we get1+2+3+...+k+(k+1)<18(2k+1)2+(k+1)⇒     1+2+3+...+k+(k+1)<18{(4k2+4k+1)+8(k+1)}⇒     1+2+3+...+k+(k+1)<18(4k2+12k+9)⇒     1+2+3+...+k+(k+1)<18(2k+3)2⇒     1+2+3+...+k+(k+1)<18{2(k+1)+1}2Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

Q.19 Prove by using principal of mathematical induction for all n ∈ N n (n + 1) (n + 5) is a multiple of 3.

Ans

Let P(n) be the given statement, i.e.,P(n): n(n+ 1) (n+ 5) is a multiple of 3Putting n=1P(1):     1(1+ 1)(1+ 5)=1.2.6        =12, which is a multiple of 3.∴P(n) is true when n=1.Let P(n) is true for some natural number k, i.e.,P(k):k(k+ 1)(k+ 5) is divisible by 3P(k):  k(k+ 1)(k+ 5)=3λ,  where λ∈N ...(1)We need to prove that P(k+1) is true whenever P(k) is true.∴(k+1){(k+1)+1}{(k+5)+1}        =(k+1)(k+2){(k+5)+1}        =(k+1)(k+2)(k+5)+(k+1)(k+2)        =k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)        =3λ+(k+1){2(k+5)+(k+2)}        =3λ+(k+1)(2k+10+k+2)        =3λ+(k+1)(3k+12)        =3λ+3(k+1)(k+4)        =3{λ+(k+1)(k+4)}        =k×p,where p={λ+(k+1)(k+4)}∈NTherefore, P(k+1) is a multiple of 3.Thus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction, P(n) is true for all n∈N.

Q.20 Prove by using principal of mathematical induction for all n ∈ N 102n – 1+ 1 is divisible by 11.

Ans

Let P(n) be the given statement, i.e., P(n):102n – 1+1 is divisible by 11Putting n=1P(1):102(1) – 1+1=10+1                =11, which is a multiple of 11.∴ P(n) is true when n=1.Let P(n) is true for some natural number k, i.e.,        P(k):102k – 1+1 is divisible by 11So,  P(k):  102k – 1+1=11λ,  where λ∈N ...(1)We need to prove that P(k+1) is true whenever P(k) is true.∴102(k+1) – 1+1=102k+2 – 1+1  =102k+1+1  =102(102k-1+1−1)+1  =102(11λ−1)+1[From equation(1)]  =1100λ−100+1  =1100λ−99  =11(100λ−9)  =11×p,where  p=(100λ−9)∈NTherefore, P(k+1) is a multiple of 11.Thus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction, P(n) is true for all n∈N.

Q.21 Prove by using principal of mathematical induction for all n ∈ N x2n – y2n is divisible by x + y.

Ans

Let P(n) be the given statement, i.e.,P(n): x2n−y2n is divisible by x+yPutting n=1P(1): x2(1)−y2(1)=x2−y2             =(x−y)(x+y), which is divisible by x+y.∴ P(n) is true when n=1.Let P(n) is true for some natural number k, i.e.,        P(k):x2k−y2k  is divisible by x+ySo,  P(k):  x2k−y2k=(x+y)λ, where  λ∈N...(1)We need to prove that P(k+1) is true whenever P(k) is true.∴ x2(k+1)−y2(k+1)=x2(k+1)−y2(k+1)   =x2k+2−y2k+2  =x2.x2k−x2.y2k+x2.y2k−y2.y2k  =x2(x2k−y2k)+y2k(x2−y2)  =x2(x+y)λ+y2k(x−y)(x+y)[From equation (1)]  =(x+y){x2λ+y2k(x−y)}  =which is a factor of (x+y)Therefore, P(k+1) is divisible by (x+y).Thus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction, P(n) is true for all n∈N.

Q.22 Prove by using principal of mathematical induction for all n ∈ N 32n+2 – 8n – 9 is divisible by 8.

Ans

Let P(n) be the given statement, i.e.,P(n):32n+2−8n−9 is divisible by 8Putting n=1P(1):32(1)+2−8(1)−9=34−8−9                     =64, which is divisible by 8∴P(n) is true when n=1.Let P(n) is true for some natural number k, i.e.,        P(k):32k+2−8k−9 is divisible by 8So,  P(k):  32k+2−8k−9=8λ,  where λ∈N...(1)We need to prove that P(k+1) is true whenever P(k) is true.∴32(k+1)+2−8(k+1)−9=32k+4−8k−8−9        =32.32k+2−8k−17        =9(8λ+8k+9)−8k−17  [From equation(1)]        =72λ+72k+81−8k−17        =72λ+64k+64        =8(9λ+8k+8)        =8×p,where  p=(9λ+8k+8)∈NTherefore, P(k+1) is divisible by 8.Thus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction, P(n) is true for all n∈N.

Q.23 Prove by using principal of mathematical induction for all n ∈ N 41n – 14n is multiple of 27.

Ans

Let P(n) be the given statement, i.e.,P(n):(41)n−(14)n is multiple of 27Putting n=1P(1):(41)1−(14)1=41−14                   =27, which is divisible by 27∴P(n) is true when n=1.Let P(n) is true for some natural number k, i.e.,        P(k):(41)k−(14)k is multiple of 27So,  P(k): (41)k−(14)k=27λ,  where λ∈N...(1)We need to prove that P(k+1) is true whenever P(k) is true.∴(41)k+1−(14)k+1=41(41)k−14(14)k        =41{27λ+(14)k}−14(14)k   [From equation(1)]        =41×27λ+41×(14)k−14(14)k        =41×27λ+(41−14)(14)k        =41×27λ+27(14)k        =27{41 λ+(14)k}        =27×p,where  p={41 λ+(14)k}∈NTherefore, P(k+1) is divisible by 27.Thus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction, P(n) is true for all n∈N.

Q.24 Prove by using principal of mathematical induction for all n ∈ N (2n + 7) < (n + 3)2.

Ans

Let P(n) be the given statement, i.e.,P(n):(2n+7)<(n+3)2P(n) is true when n=1, (2×1+7)<(1+3)2⇒               9<16Thus, P(n) is true for n=1.Let P(n) is true for some natural number k, i.e.,P(k):                (2k+7)<(k+3)2 ...(1)We need to prove that P(k+1) is true whenever P(k) is true.So,        {2(k+1)+7}<{(k+1)+3}2⇒                 (2k+9)<(k+4)2 ...(2)L.H.S.:{2(k+1)+7}=(2k+7)+2⇒    {2(k+1)+7}<(k+3)2+2 [From (1)]    {2(k+1)+7}<k2+6k+9+2    {2(k+1)+7}<k2+6k+11 ...(3)and      k2+6k+11<k2+8k+16⇒      k2+6k+11<(k+4)2 ...(4)From relation (3) and relation(4), we have {2(k+1)+7}<(k+4)2⇒    {2(k+1)+7}<{(k+1)+3}2Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer n.

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FAQs (Frequently Asked Questions)

1. Which are the topics covered in NCERT Solutions Class 11 Mathematics Chapter 4?

In NCERT Solutions Class 11 Mathematics Chapter 4, the following are the topics covered: 

  • Introduction to reasoning
  • Motivation
  • Principle of Mathematical Induction

2. Should I solve all questions in NCERT Solutions Class 11 Mathematics Chapter 4?

Students can attempt all the questions in NCERT Solutions Class 11 Mathematics Chapter 4, as there is only one exercise. It includes only 24 questions with  solutions. 

3. Are the problems in NCERT Solutions Class 11 Mathematics Chapter 4 difficult to solve?

The difficulty level in NCERT Solutions Class 11 Mathematics Chapter 4 is moderate. Therefore, students need not worry about the examples in the exercise. Besides, the solutions help the students to understand step-by-step problem solving techniques in an easy and effective manner.