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
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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
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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

GRAMMAR:-Interjection Exercises With Answers Sheet-1

Interjections An interjection is a word that expresses sudden feelings: Joy, grief, surprise, praise, anger etc. Examples: 1. Hello! How are you? 2. Hurrah! We have won the match. 3. Alas! My granny is dead! 4. Hush! The baby is asleep. 5. Hurrah! We have won the tennis match. 6. Oh! I forgot to tell you something. 7. Ah! What a pleasant surprise. 8. Oh God! I got such fright. 9. Bravo! Well done. Interjections and their Expressions: Interjection Expression Hurrah! Joy Oh dear! borrow Alas! Ah! Sorrow Ugh! disgust Oh! Wonder Good heavens! horror Interjection Expression Bravo! Praise for goodness sake! anger Fie! Hatred Hello! address My goodness! Surprise Wow! Happiness Match the following columns: Column-A Column-B Hurrah! Sorrow Oh! Happiness Hello! Wonder Wow! Joy Alas! Address Fill in the blanks: 1. _____! How can you behave like this? 2. ______! I have secured first position in the class. 3. ______! I have won a prize. 4. ______! Did you hear me? 5. _______! ...

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