NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1

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1 NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
- 1.1 Chapter 4 Principle of Mathematical Induction Exercise – 4.1

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction.

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 4
Chapter Name	Principle of Mathematical Induction
Exercise	Ex 4.1
Number of Questions Solved	24
Category	NCERT Solutions

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

NCERT Exercises

Chapter 4 Principle of Mathematical Induction Exercise – 4.1

Prove the following by using the principle of mathematical induction for aline n ∈ N :
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) : $1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ 3 }^{ n }=\frac { \left( { 3 }^{ n }-1 \right) }{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 1

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) : ${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ n }^{ 3 }={ \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 2

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) : $1+\frac { 1 }{ \left( 1+2 \right) } +\frac { 1 }{ \left( 1+2+3 \right) } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 1+2+3+.\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n \right) } =\frac { 2 }{ \left( n+1 \right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 4

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) : $1.2.3+2.3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n\left( n+1 \right) \left( n+2 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }{ 4 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 6

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) : $1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ n.3 }^{ n }=\frac { \left( 2n-1 \right) { 3 }^{ n+1 }+3 }{ 4 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 8

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) : $1.2+2.3+3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n.\left( n+1 \right) =\left[ \frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } \right]$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 10

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) : $1.3+3.5+5.7+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\left( 2n-1 \right) \left( 2n+1 \right) =\frac { n\left( { 4n }^{ 2 }+6n-1 \right) }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 12

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) : $1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n.{ 2 }^{ n }=\left( n-1 \right) { 2 }^{ n+1 }+2$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 15

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) : $\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ { 2 }^{ n } } =1-\frac { 1 }{ { 2 }^{ n } }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 17

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) : $\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 3n-1 \right) \left( 3n+2 \right) } =\frac { n }{ \left( 6n+4 \right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 18

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) : $\frac { 1 }{ 1.2.3 } +\frac { 1 }{ 2.3.4 } +\frac { 1 }{ 3.4.5 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ n\left( n+1 \right) \left( n+2 \right) } =\frac { n\left( n+3 \right) }{ 4\left( n+1 \right) \left( n+2 \right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 20

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) : $a+ar+{ ar }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ ar }^{ n-1 }=\frac { a\left( { r }^{ n }-1 \right) }{ r-1 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 22

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) : $\left( 1+\frac { 3 }{ 1 } \right) \left( 1+\frac { 5 }{ 4 } \right) \left( 1+\frac { 7 }{ 9 } \right) \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1+\frac { \left( 2n+1 \right) }{ { n }^{ 2 } } \right) ={ \left( n+1 \right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 23

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) : $\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1+\frac { 1 }{ n } \right) =\left( n+1 \right)$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 25

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) : ${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ \left( 2n-1 \right) }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 27

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) : $\frac { 1 }{ 1.4 } +\frac { 1 }{ 4.7 } +\frac { 1 }{ 7.10 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) } =\frac { n }{ \left( 3n+1 \right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 30

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) : $\frac { 1 }{ 3.5 } +\frac { 1 }{ 5.7 } +\frac { 1 }{ 7.9 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ \left( 2n+1 \right) \left( 2n+3 \right) } =\frac { n }{ 3\left( 2n+3 \right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 32

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) : $1+2+3+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n<\frac { 1 }{ 8 } { \left( 2n+1 \right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 35

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.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 36

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
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 38

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.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 39

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.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 41

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.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 42

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.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 44

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

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

Chapter 4 Principle of Mathematical Induction Exercise – 4.1

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