NCERT Solutions for Class 11 Maths Chapter 4

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Exercise 4.1 in English Medium and Hindi Medium updated for new academic session 2020-2021 based on latest new NCERT Books and following the latest CBSE Curriculum 2020-21.

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Page Contents

1 11th Maths Chapter 4 Solutions in English Medium
2 11th Maths Chapter 4 Solutions in Hindi Medium
3 11th Maths Chapter 4 Solutions in PDF
4 Important Questions on 11th Maths Chapter 4

NCERT Solutions for Class 11 Maths Chapter 4

Class: 11	Maths (English and Hindi Medium)
Chapter 4:	Principle of Mathematical Induction

11th Maths Chapter 4 Solutions

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Exercise 4.1 in English & Hindi medium is given below for new academic year 2020-2021. Visit to Class 11 Maths main page to get the solutions of all chapters. These NCERT Solutions are useful for all the students CBSE Board, Uttarakhand Board, MP Board, UP Board, Bihar Board and all other board who are following NCERT Books based on new CBSE Curriculum 2020-21.

11th Maths Chapter 4 Solutions in English Medium
- Class 11 Maths Exercise 4.1 Solutions
11th Maths Chapter 4 Solutions in Hindi Medium
- Class 11 Maths Exercise 4.1 Solution in Hindi
11th Maths Chapter 4 Solutions in PDF

Class 11 Maths Chapter 4 Exercise 4.1 Solution in Videos

Class 11 Maths Chapter 4 Exercise 4.1 Solution

Class 11 Maths Exercise 4.1 Solution in Hindi

What is meant by a statement in mathematics?

A meaningful sentence which can be judged to be either true or false is called a statement.

What is a mathematical statement?

A statement involving mathematical relations is called as mathematical statement.

What do know about Principle of mathematical Induction?

Principle of mathematical Induction:
1. Let P(n) be any statement involving natural number n such that
P(1) is true, and
2. If P(k) is true as well as P(k + 1) is true for some natural number k, i.e. P(k + 1) is true whenever P(k) is true for some natural number k, then P(n) is true for all natural numbers.

Important Terms on Mathematical Induction

Induction and deduction are two basic processes of reasoning.
1. Deduction: It is the application of a general case to a particular case. In contrast to deduction, induction is process of reasoning from particular to general.
2. Induction: Induction being with observations. From observations we arrive at tentative conclusions called conjectures. The process of induction help in proving the conjectures which may be true.

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Important Questions on 11th Maths Chapter 4

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2+〖2.2〗^2+〖3.2〗^3+⋯+n.2^n=(n-1) 2^(n+1)+2

Let the given statement be P(n), therefore,
P(n):1.2+〖2.2〗^2+〖3.2〗^3+⋯+n.2^n=(n-1) 2^(n+1)+2
For n = 1, we have
L.H.S. = 1.2 = 2
R.H.S. = (1-1) 2^(1+1)+2=0+2=2
L.H.S. = R.H.S, so P(1) is true.
Let P(k) be true for some positive integer k, such that
P(k):1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k=(k-1) 2^(k+1)+2
Now, to prove that P(k + 1) is true. i.e.
P(k+1):1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k+(k+1).2^(k+1)=(k) 2^(k+2)+2
Considering the L.H.S. of P(k + 1), we have
1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k+(k+1).2^(k+1)
=(1.2+〖2.2〗^2+〖3.2〗^3+⋯+k.2^k )+(k+1).2^(k+1)
=(k-1) 2^(k+1)+2+(k+1).2^(k+1)
[As P(k) is true]
=(k-1) 2^(k+1)+(k+1).2^(k+1)+2
=[(k-1)+(k+1)].2^(k+1)+2
=2k.2^(k+1)+2
=k.2^(k+2)+2
=R.H.S.
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.

सभी n ∈ N के लिए गणितीय आगमन सिद्धांत के प्रयोग द्वारा सिद्ध कीजिए कि: n(n+1)(n+5),संख्या 3 का एक गुणज है।

मान लीजिए कि दिया गया कथन P(n) है, अर्थात,
P(n):n(n+1)(n+5) संख्या 3 का एक गुणज है।
n = 1, के लिए,
1(1+1)(1+5)=12,जो संख्या 3 का एक गुणज है।
इसलिए, P(1) सत्य है।
माना, किसी धन पूर्णांक k के लिए, P(k) सत्य है, अर्थात
P(k):k(k+1)(k+5) संख्या 3 का एक गुणज है।
माना n(n+1)(n+5)=3m … (1)
जहाँ, m एक प्राकृतिक संख्या है।
अब सिद्ध करना है कि P(k + 1) भी सत्य है, अर्थात
P(k+1):(k+1)(k+2)(k+6) संख्या 3 का एक गुणज है।
अब, (k+1)(k+2)(k+6)
=(k+6)(k+1)(k+2)
=k(k+1)(k+2)+6(k+1)(k+2)
=3m+6(k^2+3k+2)
[समीकरण (1) से]
=3m+(6k^2+18k+12)
=3[m+(2k^2+6k+4)],जो संख्या 3 का एक गुणज है।
इस प्रकार, P(k + 1) सत्य है जब कभी P(k) सत्य है।
अतः, गणितीय आगमन सिद्धांत से सभी प्राकृत संख्याओं N के लिए कथन P(n) सत्य है।

Prove the following by using the principle of mathematical induction for all n ∈ N: 〖10〗^(2n-1)+1 is divisible by 11.

Let the given statement be P(n), therefore,
P(n):〖10〗^(2n-1)+1 is divisible by 11.
For n = 1, we have
〖10〗^(2-1)+1=11,which is divisible by 11.
So, P(1) is true.
Let P(k) be true for some positive integer k, such that
P(k):〖10〗^(2k-1)+1 is divisible by 11.
Let 〖10〗^(2n-1)+1 =11m … (1)
Where, m is any natural number.
Now, to prove that P(k + 1) is true. i.e.
P(k+1):〖10〗^(2k+1)+1 is divisible by 11.
Consider 〖10〗^(2k+1)+1
=〖10〗^(2k-1+2)+1
=〖10〗^2.〖10〗^(2k-1)+1
=〖10〗^2.(11m-1)+1
[From the equation (1),〖10〗^(2n-1)=11m-1]
=100.(11m-1)+1
=1100m – 100 + 1
=1100m – 99
=11[100m – 9],which is divisible by 11.
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.

Prove the following by using the principle of mathematical induction for all n ∈ N: x^2n-y^2n is divisible by x+y.

Let the given statement be P(n), therefore,
P(n):x^2n-y^2n is divisible by x+y.
For n = 1, we have
x^2-y^2=(x+y)(x-y),which is divisible by x+y.
So, P(1) is true.
Let P(k) be true for some positive integer k, such that
P(k):x^2k-y^2k is divisible by x+y.
Let x^2k-y^2k=(x+y)m … (1)
Where, m is any natural number.
Now, to prove that P(k + 1) is true. i.e.
P(k+1):x^2(k+1) -y^2(k+1) is divisible by x+y.
Consider x^2(k+1) -y^2(k+1)
=x^(2k+2)-y^(2k+2)
=〖x^2 x〗^2k-y^(2k+2)
=x^2 [(x+y)m+y^2k ]-y^(2k+2)
[From the equation (1),x^2k=(x+y)m+y^2k ]
=x^2 (x+y)m+x^2 y^2k-y^(2k+2)
=x^2 (x+y)m+y^2k (x^2-y^2 )
=x^2 (x+y)m+y^2k (x-y)(x+y)
=(x+y)[x^2 m+y^2k (x-y)],which is divisible by (x+y).
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.

सभी n ∈ N के लिए गणितीय आगमन सिद्धांत के प्रयोग द्वारा सिद्ध कीजिए कि: 3^(2n+2)-8n-9,संख्या 8 से भाज्य है।

मान लीजिए कि दिया गया कथन P(n) है,
अर्थात,
P(n):3^(2n+2)-8n-9 संख्या 8 से भाज्य है।
n = 1, के लिए,
3^(2+2)-8×1-9=81-17=64,संख्या 8 से भाज्य है।
इसलिए, P(1) सत्य है।
माना, किसी धन पूर्णांक k के लिए, P(k) सत्य है, अर्थात
P(k):3^(2k+2)-8k-9 संख्या 8 से भाज्य है।
माना 3^(2k+2)-8k-9 =8m … (1)
जहाँ, m एक प्राकृतिक संख्या है।
अब सिद्ध करना है कि P(k + 1) भी सत्य है, अर्थात
P(k+1):3^(2(k+1)+2)-8(k+1)-9 संख्या 8 से भाज्य है।
अब, 3^(2(k+1)+2)-8(k+1)-9
=3^(2k+4)-8k-8-9
=〖3^2 3〗^(2k+2)-8k-17
=3^2 (8m+8k+9)-8k-17
[समीकरण (1) से,
3^(2k+2)
=8m+8k+9]
=72m+72k+81-8k-17
=72m+64k+64
=8[9m+8k+8],जो संख्या 8 से भाज्य है।
इस प्रकार, P(k + 1) सत्य है जब कभी P(k) सत्य है।
अतः, गणितीय आगमन सिद्धांत से सभी प्राकृत संख्याओं N के लिए कथन P(n) सत्य है।

सभी n ∈ N के लिए गणितीय आगमन सिद्धांत के प्रयोग द्वारा सिद्ध कीजिए कि: 〖41〗^n-〖14〗^n,संख्या 27 का एक गुणज है।

मान लीजिए कि दिया गया कथन P(n) है, अर्थात,
P(n):〖41〗^n-〖14〗^n संख्या 27 का एक गुणज है।
n = 1, के लिए,
〖41〗^1-〖14〗^1=27,जो संख्या 27 का एक गुणज है।
इसलिए, P(1) सत्य है।
माना, किसी धन पूर्णांक k के लिए, P(k) सत्य है, अर्थात
P(k):〖41〗^k-〖14〗^k संख्या 27 का एक गुणज है।
माना 〖41〗^k-〖14〗^k=27m … (1)
जहाँ, m एक प्राकृतिक संख्या है।
अब सिद्ध करना है कि P(k + 1) भी सत्य है, अर्थात
P(k+1):〖41〗^(k+1)-〖14〗^(k+1) संख्या 27 का एक गुणज है।
अब, 〖41〗^(k+1)-〖14〗^(k+1)
=41.〖41〗^k-14.〖14〗^k
=41.(27m+〖14〗^k )-14.〖14〗^k
[समीकरण (1) से,〖41〗^k=27m+〖14〗^k ]
=41×27m+41.〖14〗^k-14.〖14〗^k
=41×27m+27.〖14〗^k
=27[41m+〖14〗^k ],जो संख्या 27 का एक गुणज है।
इस प्रकार, P(k + 1) सत्य है जब कभी P(k) सत्य है।
अतः, गणितीय आगमन सिद्धांत से सभी प्राकृत संख्याओं N के लिए कथन P(n) सत्य है।

Prove the following by using the principle of mathematical induction for all n ∈ N: (2n+7)<(n+3)^2.

Let the given statement be P(n),
therefore,
P(n):(2n+7)<(n+3)^2. For n = 1, we have (2×1+7)<(1+3)^2 ⇒9<16 ⇒ P(1) is true. Let P(k) be true for some positive integer k, such that P(k):(2k+7)<(k+3)^2. Now, to prove that P(k + 1) is true. i.e. P(k+1):(2k+9)<(k+4)^2 Considering the statement P(k), we have (2k+7)<(k+3)^2 [Difference between (k+4)^2 and (k+3)^2=(k+4)^2-(k+3)^2=k^2+8k+16-(k^2+6k+9)=2k+7] Adding both sides (2k+7), we have (2k+7)+(2k+7)<(k+3)^2+(2k+7) ⇒(2k+9)+(2k+5)<(k^2+6k+9)+(2k+7) ⇒(2k+9)+(2k+5)<(k^2+6k+9+2k+7) ⇒(2k+9)+(2k+5)<(k^2+8k+16) ⇒(2k+9)+(2k+5)<(k+4)^2 ⇒(2k+9)<(k+4)^2 [As k>0,so 2k+1>0]
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.