Class 11 Maths Chapter 4 Principle of Mathematical Induction

Exercise – 4.1

Prove the following by using the principle of mathematical induction for a line 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
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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) : $1.2.3+2.3.4+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +n\left( n+1 \right) \left( n+2 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }{ 4 }$
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 6
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 7

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

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
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 13

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) : $\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ { 2 }^{ n } } =1-\frac { 1 }{ { 2 }^{ n } }$
tiwari academy 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) : $\frac { 1 }{ 1.2.3 } +\frac { 1 }{ 2.3.4 } +\frac { 1 }{ 3.4.5 } +\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +\frac { 1 }{ n\left( n+1 \right) \left( n+2 \right) } =\frac { n\left( n+3 \right) }{ 4\left( n+1 \right) \left( n+2 \right) }$
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 20
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 21

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
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 24

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) : ${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot +{ \left( 2n-1 \right) }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 }$
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 27
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 28

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
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 31

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
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 33

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
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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.
tiwari academy 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.
tiwari academy class 11 maths Chapter 4 Principle of Mathematical Induction 44

VERB AND FORMS OF VERB WITH HINDI MEANING_List of Verb in hindi

Verbs What is a verb? Verbs are the action words in a sentence that describe what the subject is doing. Along with nouns, verbs are the main part of a sentence or phrase, telling a story about what is taking place. Verbs are words that express action or state of being. There are three types of verbs: action verbs, linking verbs, and helping verbs . Action verbs are words that express action (give, eat, walk, etc.) or possession (have, own, etc.). Action verbs can be either transitive or intransitive. List of Verb in hindi Present Hindi Meaning Past Past Participle Buy खरीदना Bought Bought Build बांधना Built Built burn जलना Burnt Burnt Bend झुकना Bent Bent Bring लाना brought brought Become होना Became Become Come आना Came Come Catch पकड़ना Caught Caught Do करना Did Done Dream ख्वाब देखना Dreamt Dreamt Arise उठना / जागना Arose Arisen Be होना Was, were Been Bear सहन करना Bore Bore Beat मारना Beat Beat Bite काटना Bit Bitten Break तोडना Broke Broken Choose चुनना Chose Chosen Draw चि

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