NCERT Solutions for Class 11 Maths Chapter 4 is given by people having the right knowledge and subject expertise. After solving the NCERT Solutions for 11th Class Maths Principle of Mathematical Induction, students can score well in the board exams. You can understand the topics very clearly with the accurate and easily understandable NCERT Solutions for Class 11 Maths Ch 4. Furthermore, you can download the CBSE Class 11 NCERT Maths Solutions for Principle of Mathematical Induction through the quick links prevailing.
Get to know about NCERT Solutions for Class 11 Maths Chapter 4 Ex 4.1, Miscellaneous Exercise in PDF Format. To help you out we have jotted down the Class 11 Maths NCERT Solutions of Chapter 4 in Hindi & English Mediums to meet your requirements. Students of various boards such as UP, MP, Uttarakhand, Gujarat, and many others can refer to these Class 11 Maths NCERT Solutions and score well.
Class 11 Maths NCERT Solutions Chapter 4 Principle of Mathematical Induction
Students of Class 11 Maths can learn about the Principle of Mathematical Induction and its application in detail. It is a Specific Technique to prove few mathematically accepted statements in algebra and in other applications of Mathematics like inductive and deductive reasoning. In this Chapter of Class 11th Maths, you will learn how to prove a formula and how an equation is derived. NCERT Solutions of Class 11 Maths Ch 4 covers all the concepts so that you can crack any question from the Principle of Mathematical Induction. Exercise wise 11th Class Maths NCERT Solutions for Chapter 4 are very accurate and make it easy for you to score good marks in the exam.
Make the most out of the NCERT Class 11 Maths Chapter 4 Solutions for Ex 4.1, Miscellaneous Exercise by either viewing or downloading them through the direct links available.
| Class |
11 |
| Book |
Mathematics |
| Subject |
Maths |
| Chapter Number |
4 |
| Name of the Chapter |
Principle of Mathematical Induction |
NCERT Solutions of Class 11 Maths Ch 4 – Solved Exercises
Class 11 Maths NCERT Solutions for Chapter 4 Principle of Mathematical Induction is provided here. With the handy study material, you can easily score good marks as all the 11th Class Maths NCERT Solutions of Ch 4 follow a step by step approach. In addition to the standard NCERT Solutions, you can also get information concerning Notes, Exemplar Problems, Important Questions, Formula. Download the preparation related stuff via quick links available for free of cost.
Chapter 4 Principle of Mathematical Induction Exercise – 4.1
Prove the following by using the principle of mathematical induction for aline n ∈ N :
Ex 4.1 Class 11 Maths Question 1.
\(1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ 3 }^{ n }=\frac { \left( { 3 }^{ n }-1 \right) }{ 2 } \)
Solution.
Let the given statement be P(n) i.e.,
P(n) : \(1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ 3 }^{ n }=\frac { \left( { 3 }^{ n }-1 \right) }{ 2 } \)
Ex 4.1 Class 11 Maths Question 2.
\({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ n }^{ 3 }={ \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }\)
Solution.
Let the given statement be P(n) i.e.,
P(n) : \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ n }^{ 3 }={ \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }\)
Ex 4.1 Class 11 Maths Question 3.
\(1+\frac { 1 }{ \left( 1+2 \right) } +\frac { 1 }{ \left( 1+2+3 \right) } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 1+2+3+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n \right) } =\frac { 2 }{ \left( n+1 \right) } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1+\frac { 1 }{ \left( 1+2 \right) } +\frac { 1 }{ \left( 1+2+3 \right) } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 1+2+3+.\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n \right) } =\frac { 2 }{ \left( n+1 \right) } \)
Ex 4.1 Class 11 Maths Question 4.
\(1.2.3+2.3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n\left( n+1 \right) \left( n+2 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }{ 4 } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1.2.3+2.3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n\left( n+1 \right) \left( n+2 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }{ 4 } \)
Ex 4.1 Class 11 Maths Question 5.
\(1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ n.3 }^{ n }=\frac { \left( 2n-1 \right) { 3 }^{ n+1 }+3 }{ 4 } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ n.3 }^{ n }=\frac { \left( 2n-1 \right) { 3 }^{ n+1 }+3 }{ 4 } \)
Ex 4.1 Class 11 Maths Question 6.
\(1.2+2.3+3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n.\left( n+1 \right) =\left[ \frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } \right] \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1.2+2.3+3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n.\left( n+1 \right) =\left[ \frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } \right] \)
Ex 4.1 Class 11 Maths Question 7.
\(1.3+3.5+5.7+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\left( 2n-1 \right) \left( 2n+1 \right) =\frac { n\left( { 4n }^{ 2 }+6n-1 \right) }{ 3 } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1.3+3.5+5.7+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\left( 2n-1 \right) \left( 2n+1 \right) =\frac { n\left( { 4n }^{ 2 }+6n-1 \right) }{ 3 } \)
Ex 4.1 Class 11 Maths Question 8.
\(1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n.{ 2 }^{ n }=\left( n-1 \right) { 2 }^{ n+1 }+2\)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n.{ 2 }^{ n }=\left( n-1 \right) { 2 }^{ n+1 }+2\)
Ex 4.1 Class 11 Maths Question 9
\(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ { 2 }^{ n } } =1-\frac { 1 }{ { 2 }^{ n } } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ { 2 }^{ n } } =1-\frac { 1 }{ { 2 }^{ n } } \)
Ex 4.1 Class 11 Maths Question 10.
\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 3n-1 \right) \left( 3n+2 \right) } =\frac { n }{ \left( 6n+4 \right) } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 3n-1 \right) \left( 3n+2 \right) } =\frac { n }{ \left( 6n+4 \right) } \)
Ex 4.1 Class 11 Maths Question 11.
\(\frac { 1 }{ 1.2.3 } +\frac { 1 }{ 2.3.4 } +\frac { 1 }{ 3.4.5 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ n\left( n+1 \right) \left( n+2 \right) } =\frac { n\left( n+3 \right) }{ 4\left( n+1 \right) \left( n+2 \right) } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\frac { 1 }{ 1.2.3 } +\frac { 1 }{ 2.3.4 } +\frac { 1 }{ 3.4.5 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ n\left( n+1 \right) \left( n+2 \right) } =\frac { n\left( n+3 \right) }{ 4\left( n+1 \right) \left( n+2 \right) } \)
Ex 4.1 Class 11 Maths Question 12.
\(a+ar+{ ar }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ ar }^{ n-1 }=\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(a+ar+{ ar }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ ar }^{ n-1 }=\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } \)
Ex 4.1 Class 11 Maths Question 13.
\(\left( 1+\frac { 3 }{ 1 } \right) \left( 1+\frac { 5 }{ 4 } \right) \left( 1+\frac { 7 }{ 9 } \right) \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1+\frac { \left( 2n+1 \right) }{ { n }^{ 2 } } \right) ={ \left( n+1 \right) }^{ 2 }\)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\left( 1+\frac { 3 }{ 1 } \right) \left( 1+\frac { 5 }{ 4 } \right) \left( 1+\frac { 7 }{ 9 } \right) \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1+\frac { \left( 2n+1 \right) }{ { n }^{ 2 } } \right) ={ \left( n+1 \right) }^{ 2 }\)
Ex 4.1 Class 11 Maths Question 14.
\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1+\frac { 1 }{ n } \right) =\left( n+1 \right) \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1+\frac { 1 }{ n } \right) =\left( n+1 \right) \)
Ex 4.1 Class 11 Maths Question 15.
\({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ \left( 2n-1 \right) }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ \left( 2n-1 \right) }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 } \)
Ex 4.1 Class 11 Maths Question 16.
\(\frac { 1 }{ 1.4 } +\frac { 1 }{ 4.7 } +\frac { 1 }{ 7.10 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) } =\frac { n }{ \left( 3n+1 \right) } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\frac { 1 }{ 1.4 } +\frac { 1 }{ 4.7 } +\frac { 1 }{ 7.10 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) } =\frac { n }{ \left( 3n+1 \right) } \)
Ex 4.1 Class 11 Maths Question 17.
\(\frac { 1 }{ 3.5 } +\frac { 1 }{ 5.7 } +\frac { 1 }{ 7.9 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 2n+1 \right) \left( 2n+3 \right) } =\frac { n }{ 3\left( 2n+3 \right) } \)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(\frac { 1 }{ 3.5 } +\frac { 1 }{ 5.7 } +\frac { 1 }{ 7.9 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 2n+1 \right) \left( 2n+3 \right) } =\frac { n }{ 3\left( 2n+3 \right) } \)
Ex 4.1 Class 11 Maths Question 18.
\(1+2+3+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n<\frac { 1 }{ 8 } { \left( 2n+1 \right) }^{ 2 }\)
Solution.
Let the given statement be P(n), i.e.,
P(n) : \(1+2+3+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n<\frac { 1 }{ 8 } { \left( 2n+1 \right) }^{ 2 }\)
Ex 4.1 Class 11 Maths Question 19.
n(n+1 )(n + 5) is a multiple of 3.
Solution.
Let the given statement be P(n), i.e.,
P(n): n(n + l)(n + 5) is a multiple of 3.
Ex 4.1 Class 11 Maths Question 20.
\({ 10 }^{ 2n-1 }+1\) is divisible by 11.
Solution.
Let the given statement be P(n), i.e.,
P(n): \({ 10 }^{ 2n-1 }+1\) is divisible by 11
Ex 4.1 Class 11 Maths Question 21.
\({ x }^{ 2n }-{ y }^{ 2n }\) is divisible by x + y.
Solution.
Let the given statement be P(n), i.e.,
P(n): \({ x }^{ 2n }-{ y }^{ 2n }\) is divisible by x + y.
Ex 4.1 Class 11 Maths Question 22.
\({ 3 }^{ 2n+2 }-8n-9\) is divisible by 8.
Solution.
Let the given statement be P(n), i.e.,
P(n): \({ 3 }^{ 2n+2 }-8n-9\) is divisible by 8.
Ex 4.1 Class 11 Maths Question 23.
\({ 41 }^{ n }-{ 14 }^{ n }\) is a multiple of 27.
Solution.
Let the given statement be P(n), i.e.,
P(n): \({ 41 }^{ n }-{ 14 }^{ n }\) is a multiple of 27.
Ex 4.1 Class 11 Maths Question 24.
\(\left( 2n+7 \right) <{ \left( n+3 \right) }^{ 2 }\)
Solution.
Let the given statement be P(n), i.e.,
P(n): \(\left( 2n+7 \right) <{ \left( n+3 \right) }^{ 2 }\)
First we prove that the statement is true for n = 1.
Make the most out of the Class 11 Maths Chapter 4 Principle of Mathematical Induction Exercise 4.1, Miscellaneous Exercise Solutions, and crack the exam with great scores. Keeping in mind the need of aspirants like you we have curated the CBSE NCERT Solutions for 11th Class Maths Chapter 4 PDF in Hindi & English Mediums. You can view them as per your convenience and get a good grasp of the concepts involved in them.
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