NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction – Here are all the NCERT solutions for Class 11 Maths Chapter 4. This solution contains questions, answers, images, explanations of the complete chapter 4 titled Of Principle of Mathematical Induction taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 4 Principle of Mathematical Induction After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction in one place.
NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 4 Principle of Mathematical Induction , Maths, Class 11.
| Class | 11 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 4 |
| Chapter Name |
Principle of Mathematical Induction |
NCERT Solutions Class 11 Maths chapter 4 Principle of Mathematical Induction
Class 11, Maths chapter 4, Principle of Mathematical Induction solutions are given below in PDF format. You can view them online or download PDF file for future use.
Principle of Mathematical Induction
Did you find NCERT Solutions Class 11 Maths chapter 4 Principle of Mathematical Induction helpful? If yes, please comment below. Also please like, and share it with your friends!
NCERT Solutions Class 11 Maths chapter 4 Principle of Mathematical Induction- Video
You can also watch the video solutions of NCERT Class11 Maths chapter 4 Principle of Mathematical Induction here.
Video – will be available soon.
If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.
Download NCERT Solutions Class 11 Maths chapter 4 Principle of Mathematical Induction In PDF Format
You can also download here the NCERT Solutions Class 11 Maths chapter 4 Principle of Mathematical Induction in PDF format.
Click Here to download NCERT Solutions for Class 11 Maths chapter 4 Principle of Mathematical Induction
Question & Answer
Q.1: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1+3+3^{2}+\ldots+3^{a-1}=\frac{\left(3^{n}-1\right)}{2}\)
Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {P(n) : 1+3+3^{2}+\ldots+3^{n-1}=\frac{\left(3^{n}-1\right)}{2}} \\ {\text { For } n=1, \text { we have }} \\ {P(1) : 1=\frac{\left(3^{\prime}-1\right)}{2}=\frac{3-1}{2}=\frac{2}{2}=1}\end{array}\) , which is true \(\begin{array}{l}{\text { Let } P(k) \text { be true for some positive integer } k, \text { i.e., }} \\ {1+3+3^{2}+\ldots+3^{k-1}=\frac{\left(3^{k}-1\right)}{2}} \\ {\text { We shall now prove that } P(k+1) \text { is true. }} \\ {\text { Consider }} \\ {1+3+3^{2}+\ldots+3^{k-1}+3^{(k+1)-1}} \\ {=\left(1+3+3^{2}+\ldots+3^{k-1}\right)+3^{k}}\end{array}\) \(\begin{array}{l}{=\frac{\left(3^{t}-1\right)}{2}+3^{k}} \\ {=\frac{\left(3^{t}-1\right)+2.3^{k}}{2}}\end{array}\) [using (i)] \(=\frac{(1+2) 3^{k}-1}{2}\) \(\begin{aligned} &=\frac{3.3^{k}-1}{2} \\ &=\frac{3^{k+1}-1}{2} \end{aligned}\) \(\begin{array}{l}{\text { Thus, } P(k+1) \text { is true whenever } P(k) \text { is true. }} \\ {\text { Hence, by the principle of mathematical induction, statement } P(n) \text { is true for all natural }} \\ {\text { numbers l.e., }}\end{array}\)
Q.2: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}\)
Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {\quad 1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}} \\ {\text { For } n=1, \text { we have }} \\ {\text { For } n=1, \text { we have }} \\ {P(1) : 1^{3}=1=\left( \begin{array}{c}{1(1+1)} \\ \hline\end{array}\right.^{2}=\left(\frac{1.2}{2}\right)^{2}=1^{2}=1}\end{array}\) , which is true. \(\begin{array}{l}{\text { Let } P(k) \text { be true for some positive integer } k, \text { l.e., }} \\ {1^{3}+2^{3}+3^{3}+\ldots \ldots+k^{3}=\left(\frac{k(k+1)}{2}\right)^{2}} \\ {\text { We shall now prove that } P(k+1) \text { is true. }} \\ {\text { Consider }} \\ {1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3}}\end{array}\) \(=\left(\frac{k(k+1)}{2}\right)^{2}+(k+1)^{3} \quad[\text { Using }(i)]\) \(\begin{aligned} &=\frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3} \\ &=\frac{k^{2}(k+1)^{2}+4(k+1)^{3}}{4} \\ &=\frac{(k+1)^{2}\left\{k^{2}+4(k+1)\right\}}{4} \end{aligned}\) \(\begin{array}{l}{=\frac{(k+1)^{2}\left\{k^{2}+4 k+4\right\}}{4}} \\ {=\frac{(k+1)^{2}(k+2)^{2}}{4}} \\ {=\frac{(k+1)^{2}(k+1+1)^{2}}{4}}\end{array}\) \(=\left(1^{3}+2^{3}+3^{3}+\ldots .+k^{3}\right)+(k+1)^{3}=\left(\frac{(k+1)(k+1+1)}{2}\right)^{2}\) \(\begin{array}{l}{\text { Thus, } P(k+1) \text { is true whenever } P(k) \text { is true. }} \\ {\text { Hence, by the principle of mathematical induction, statement } P(n) \text { is true for all natural }} \\ {\text { numbers l.e., }}\end{array}\)
Q.3: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots+\frac{1}{(1+2+3+\ldots n)}=\frac{2 n}{(n+1)}\)
Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {\text { P(n): } 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots n}=\frac{2 n}{n+1}} \\ {\text { For } n=1, \text { we have }} \\ {P(1) : 1=\frac{2.1}{1+1}=\frac{2}{2}=1} \\ {\text { Let } P(k) \text { be true for some positive integer } k, \text { i.e., }}\end{array}\) \(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}=\frac{2 k}{k+1}\)............(i) \(\begin{array}{l}{\text { We shall now prove that } \mathrm{P}(k+1) \text { is true. }} \\ {\text { Consider }} \\ {1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}+\frac{1}{1+2+2+k+(k+1)}}\end{array}\) \(\begin{array}{l}{=\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+. k}\right)+\frac{1}{1+2+3+\ldots+k+(k+1)}} \\ {=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots+k+(k+1)} \quad \quad[\text { Using (i) }]}\end{array}\) \(=\frac{2 k}{k+1}+\frac{1}{\left(\frac{(k+1)(k+1+1)}{2}\right)}\) \(\left[1+2+3+\ldots+n=\frac{n(n+1)}{2}\right]\) \(\begin{array}{l}{=\frac{2 k}{(k+1)}+\frac{2}{(k+1)(k+2)}} \\ {=\frac{2}{(k+1)}\left(k+\frac{1}{k+2}\right)}\end{array}\)
Q.4: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1.2.3+2.3 .4+\ldots+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}\)
Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {P(n) : 1.2 .3+2.3 .4+\ldots+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}}\end{array}\) \(\begin{array}{l}{\text { For } n=1, \text { we have }} \\ {P(1) : 1.2 \cdot 3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.23 .4}{4}=6}\end{array}\) , which is true Let P(k) be true for some positive integer k, i.e., \(1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}\) ……………(i) \(\begin{array}{l}{\text { We shall now prove that } P(k+1) \text { is true. }} \\ {\text { Consider }} \\ {1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)+(k+1)(k+2)(k+3)} \\ {=\{1.2 .3+2.3 .4+\ldots+k(k+1)(k+2)\}+(k+1)(k+2)(k+3)}\end{array}\) \(\begin{array}{l}{=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3) \quad[\text { Using }(i)]} \\ {=(k+1)(k+2)(k+3)\left(\frac{k}{4}+1\right)} \\ {=\frac{(k+1)(k+2)(k+3)(k+4)}{4}} \\ {=\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}}\end{array}\) \(\begin{array}{l}{\text { Thus, } P(k+1) \text { is true whenever } P(k) \text { is true. }} \\ {\text { Hence, by the principle of mathematical induction, statement } P(n) \text { is true for all natural }} \\ {\text { numbers l.e., } n \text { . }}\end{array}\)
Q.5: Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1.3+2.3^{2}+3.3^{3}+\ldots+n 3^{n}=\frac{(2 n-1) 3^{n+1}+3}{4}\)
Ans : \(\begin{array}{l}{\text { Let the given statement be } P(n), \text { i.e., }} \\ {1.3+2.3^{2}+3.3^{\circ}+\ldots+n 3^{n}=\frac{(2 n-1) 3^{n+1}+3}{4}}\end{array}\) \(\begin{array}{l}{P(n) :} \\ {\text { For } n=1, \text { we have }}\end{array}\) \(P(1) : 1.3=3 \quad=\frac{(2.1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}=3\) \(\begin{array}{l}{\text { Let } P(k) \text { be true for some positive integer } k, \text { i.e., }} \\ {1.3+2.3^{2}+3.3^{3}+\ldots+k 3^{4}=\frac{(2 k-1) 3^{k+1}+3}{4}} \\ {\text { We shall now prove that } P(k+1) \text { is true. }}\end{array}\) ……….(i) Consider \(\begin{array}{l}{1.3+2.3^{2}+3.3^{3}+\ldots+k 3^{k}+(k+1) 3^{k+1}} \\ {=\left(1.3+2.3^{2}+3.3^{3}+\ldots+k .3^{k}\right)+(k+1) 3^{k+1}}\end{array}\) \(=\frac{(2 k-1) 3^{k+1}+3}{4}+(k+1) 3^{k+1} \quad[\text { Using }(\mathbf{i})]\) \(\begin{array}{l}{=\frac{(2 k-1) 3^{3+1}+3+4(k+1) 3^{k+1}}{4}} \\ {=\frac{3^{t+1}\{2 k-1+4(k+1)\}+3}{4}}\end{array}\) \(\begin{aligned} &=\frac{3^{k+1}\{6 k+3\}+3}{4} \\ &=\frac{3^{k+1} \cdot 3\{2 k+1\}+3}{4} \end{aligned}\) \(\begin{array}{l}{=\frac{3^{(k+1)+1}\{2 k+1\}+3}{4}} \\ {=\frac{\{2(k+1)-1\} 3^{(k+1)+1}+3}{4}}\end{array}\) 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 i.e., n.
NCERT / CBSE Book for Class 11 Maths
You can download the NCERT Book for Class 11 Maths in PDF format for free. Otherwise you can also buy it easily online.
- Click here for NCERT Book for Class 11 Maths
- Click here to buy NCERT Book for Class 11 Maths
All NCERT Solutions Class 11
- NCERT Solutions for Class 11 Accountancy
- NCERT Solutions for Class 11 Biology
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Economics
- NCERT Solutions for Class 11 History
- NCERT Solutions for Class 11 Geography
- NCERT Solutions for Class 11 Political Science
- NCERT Solutions for Class 11 Sociology
- NCERT Solutions for Class 11 Psychology
- NCERT Solutions for Class 11 English
- NCERT Solutions for Class 11 Hindi
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Business Studies
- NCERT Solutions for Class 11 Statistics
All NCERT Solutions
You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
| Class 4 | Class 5 | Class 6 |
| Class 7 | Class 8 | Class 9 |
| Class 10 | Class 11 | Class 12 |
Download the NCERT Solutions app for quick access to NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction. It will help you stay updated with relevant study material to help you top your class!
The post NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction appeared first on AglaSem Schools.
from AglaSem Schools https://ift.tt/38e1nVh
https://ift.tt/eA8V8J https://ift.tt/eA8V8J
Post a Comment