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

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

Edited By Vishal kumar | Updated on Nov 03, 2023 10:56 AM IST

NCERT Solutions for Class 11 Maths Chapter 4 - Principle of Mathematical Induction Exercise 4.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction Exercise 4.1- NCERT solutions for exercise 4.1 Class 11 Maths Chapter 4 deals with the principle of mathematical induction and related problems. Exercise 4.1 Class 11 Maths deals with how to prove a given mathematical statement using the principle of mathematical induction. NCERT Solutions for Class 11 Maths chapter 4 exercise 4.1 gives an insight into the steps of proving a given statement using the idea of induction. Solving both example questions and Class 11 Maths chapter 4 exercise 4.1 are important to understand the concepts discussed in the chapter. Before starting the Class 11 Maths chapter 4 exercise 4.1 the NCERT book discusses the concept of deduction in brief and introduces what is induction and how induction is different from deduction. Followed by examples and Class 11 Maths chapter 4 exercise 4.1.

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  4. Question:1 Prove the following by using the principle of mathematical induction for all :
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NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction
NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

The NCERT Solutions for Class 11 Maths Exercise 4.1 are expertly crafted to help students overcome challenges in problem-solving. They provide step-by-step explanations and a free PDF version for convenient access, ensuring that students can grasp mathematical concepts effectively and excel in their studies.

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

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Access Principle Of Mathematical Induction Class 11 Chapter 4- Exercise 4.1

Question:1 Prove the following by using the principle of mathematical induction for all n\in N : 1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}

Answer:

Let the given statement be p(n) i.e.
p(n):1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}
For n = 1 we have
p(1): 1=\frac{(3^1-1)}{2}=\frac{3-1}{2}= \frac{2}{2}=1 , which is true

For n = k we have
p(k):1+3+3^2+...+3^{k-1}=\frac{(3^k-1)}{2} \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1+3+3^2+...+3^{k+1-1}= 1+3+3^2+...+3^{k-1}+3^{k}
= (1+3+3^2+...+3^{k-1})+3^{k}
= \frac{(3^k-1)}{2}+3^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Using \ (i))
= \frac{3^k-1+2.3^k}{2}
= \frac{3^k(1+2)-1}{2}
= \frac{3.3^k-1}{2}
= \frac{3^{k+1}-1}{2}
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 n

Question:2 Prove the following by using the principle of mathematical induction for all n\in N : 1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2

Answer:

Let the given statement be p(n) i.e.
p(n):1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2
For n = 1 we have
p(1):1=\left (\frac{1(1+1)}{2} \right )^2= \left ( \frac{1(2)}{2} \right )^2=(1)^2=1 , which is true

For n = k we have
p(k):1^3+2^3+3^3+...+k^3=\left (\frac{k(k+1)}{2} \right )^2 \ \ \ \ \ \ \ \ \ \ - (i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1^3+2^3+3^3+...+(k+1)^3=1^3+2^3+3^3+...+k^3+(k+1)^3
=(1^3+2^3+3^3+...+k^3)+(k+1)^3
=\left ( \frac{k(k+1)}{2} \right )^2+(k+1)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k^2(k+1)^2+4(k+1)^3}{4}
=\frac{(k+1)^2(k^2+4(k+1))}{4}
=\frac{(k+1)^2(k^2+4k+4)}{4}
=\frac{(k+1)^2(k+2)^2}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a^2+2ab+b^2=(a+b)^2)
=\left ( \frac{(k+1)(k+2)}{2} \right )^2
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 n

Question:3 Prove the following by using the principle of mathematical induction for all n\in N :

1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}

Answer:

Let the given statement be p(n) i.e.
p(n):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}
For n = 1 we have
p(1):1=\left (\frac{2(1)}{1+1} \right )= \left ( \frac{2}{2} \right )=1 , which is true

For n = k we have
p(k):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)}=\frac{2k}{(k+1)} \ \ \ \ -(i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k+1)} =\left ( 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)} \right )+\frac{1}{(1+2+3+...+k+k+1)}

=\frac{2k}{k+1}+\frac{1}{(1+2+3+...+k+(k+1))} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+1+1)}{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because1+2+....+n = \frac{n(n+1)}{2} )
=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}
=\frac{2}{k+1}\left (k+\frac{1}{k+2} \right )
=\frac{2}{k+1}\left ( \frac{k^2+2k+1}{k+2} \right )
=\frac{2}{k+1}.\frac{(k+1)^2}{k+2}
=\frac{2(k+1)}{k+2}
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 n

Question:4 Prove the following by using the principle of mathematical induction for all n\in N : 1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}

Answer:

Let the given statement be p(n) i.e.
p(n):1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}
For n = 1 we have
p(1):6=\left (\frac{1(1+1)(1+2)(1+3)}{4} \right )= \left ( \frac{1.2.3.4}{4} \right )=6 , which is true

For n = k we have
p(k):1.2.3+2.3.4+...+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4} \ \ \ \ \ \ -(i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1.2.3+2.3.4+...+k(k+1)(k+2) + (k+1)(k+2)(k+3) =(1.2.3+2.3.4+...+k(k+1)(k+2)) + (k+1)(k+2)(k+3)

=\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3) }{4}
=\frac{(k+1)(k+2)(k+3)(k+4) }{4}
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 n

Question:5 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.3+2.3^2+3.3^3+...+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}

Answer:

Let the given statement be p(n) i.e.
p(n):1.3+2.3^2+3.3^3+...+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}
For n = 1 we have
p(1):3=\frac{(2(1)-1)3^{1+1}+3}{4}= \frac{(2-1)9+3}{4}=\frac{12}{4}=3 , which is true

For n = k we have

p(k):1.3+2.3^2+3.3^3+...+k.3^k=\frac{(2k-1)3^{k+1}+3}{4} \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1.3+2.3^2+3.3^3+...+(k+1).3^{(k+1)}=1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{(k+1)}

=\frac{(2k-1)3^{k+1}+3}{4}+(k+1).3^{(k+1)}
=\frac{(2k-1)3^{k+1}+3+4(k+1).3^{(k+1)}}{4}
=\frac{3^{k+1}((2k-1)+4(k+1))+3}{4}
=\frac{3^{k+1}(6k+3)+3}{4}
=\frac{3^{k+1}.3(2k+1)+3}{4}
=\frac{(2k+1)3^{k+2}+3}{4}
=\frac{(2(k+1)-1)3^{(k+1)+1}+3}{4}
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 n

Question:6 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.2+2.3+3.4+...+n.(n+1)=\left [\frac{n(n+1)(n+2)}{3} \right ]

Answer:

Let the given statement be p(n) i.e.
p(n):1.2+2.3+3.4+...+n.(n+1)=\left [\frac{n(n+1)(n+2)}{3} \right ]
For n = 1 we have
p(1):2=\left [\frac{1(1+1)(1+2)}{3} \right ]= \frac{1.2.3}{3}=2 , which is true

For n = k we have

p(k):1.2+2.3+3.4+...+k.(k+1)=\left [\frac{k(k+1)(k+2)}{3} \right ] \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1.2+2.3+3.4+...+(k+1).(k+2)=1.2+2.3+3.4+...+k(k+1)+(k+1).(k+2)

=\frac{k(k+1)(k+2)}{3}+(k+1).(k+2) \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(k+1)(k+2)+3(k+1).(k+2)}{3}
=\frac{(k+1)(k+2)(k+3)}{3}
=\frac{(k+1)(k+1+1)(k+1+2)}{3}

Thus, p(k+1) is true whenever p(k) is true
Hence, by principle of mathematical induction , statement p(n) is true for all natural numbers n

Question:7 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}

Answer:

Let the given statement be p(n) i.e.
p(n):1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}
For n = 1 we have
p(1):1.3=3=\frac{1(4(1)^2+6(1)-1)}{3}= \frac{4+6-1}{3}=\frac{9}{3}=3 , which is true

For n = k we have

p(k):1.3+3.5+5.7+...+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3} \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1.3+3.5+5.7+...+(2(k+1)-1)(2(k+1)+1)=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)

=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3) \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(4k^2+6k-1)+3(2k+1)(2k+3)}{3}
=\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}
=\frac{(4k^3+6k^2-k+12k^2+28k+9)}{3}
=\frac{(4k^3+18k^2+23k+9)}{3}
=\frac{(4k^3+14k^2+9k+4k^2+14k+9)}{3}
=\frac{(k(4k^2+14k+9)+4k^2+14k+9)}{3}
=\frac{(4k^2+14k+9)(k+1)}{3}
=\frac{(k+1)(4k^2+8k+4+6k+6-1)}{3}
=\frac{(k+1)(4(k^2+2k+1)+6(k+1)-1)}{3}
=\frac{(k+1)(4(k+1)^2+6(k+1)-1)}{3}

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 n

Question:8 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.2+2.2^2+3.2^3+....+n.2^n=(n-1)n^{n+1}+2

Answer:

Let the given statement be p(n) i.e.
p(n):1.2+2.2^2+3.2^3+...+n.2^n=(n-1)2^{n+1}+2
For n = 1 we havep(k):1.2+2.2^2+3.2^3+...+k.2^k=(k-1)2^{k^+1}+2 \ \ \ \ \ \ -(i)p(1):1.2=2=(1-1)2^{1^+1}+2 = 2 , which is true

For n = k we have

p(k):1.2+2.2^2+3.2^3+...+k.2^k=(k-1)2^k^+^1+2 \ \ \ \ \ \ -(i)

, Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1.2+2.2^2+3.2^3+...+(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} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=2^{k+1}(k-1+k+1)+2
=2^{k+1}(2k)+2
=k.2^{k+2}+2
=(k+1-1).2^{k+1+1}+2

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 n

Question:9 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}
For n = 1 we have
p(1):\frac{1}{2}=1-\frac{1}{2^1}= 1-\frac{1}{2} = \frac{1}{2} , which is true

For n = k we have

p(k):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}=1-\frac{1}{2^k} \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{k+1}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}+\frac{1}{2^{k+1}}

=1-\frac{1}{2^k}+\frac{1}{2^{k+1}} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=1-\frac{1}{2^k}\left (1-\frac{1}{2} \right )
=1-\frac{1}{2^k}\left (\frac{1}{2} \right )
=1-\frac{1}{2^{k+1}}

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 n

Question:10 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}
For n = 1 we have
p(1):\frac{1}{2.5}= \frac{1}{10}=\frac{1}{(6(1)+4)}= \frac{1}{10} , which is true

For n = k we have

p(k):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)} \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3(k+1)-1)(3(k+1)+2)}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)(3(k+1)+2)}

=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{1}{3k+2}\left ( \frac{k}{2}+\frac{1}{3k+5} \right )
=\frac{1}{3k+2}\left ( \frac{k(3k+5)+2}{2(3k+5)} \right )
=\frac{1}{3k+2}\left ( \frac{3k^2+5k+2}{2(3k+5)} \right )
=\frac{1}{3k+2}\left ( \frac{3k^2+3k+2k+2}{2(3k+5)} \right )
=\frac{1}{3k+2}\left ( \frac{(3k+2)(k+1)}{2(3k+5)} \right )
=\frac{(k+1)}{6k+10}
=\frac{(k+1)}{6(k+1)+4}

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 n

Question:11 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}

Answer:

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}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}
For n = 1 we have
p(1):\frac{1}{1.2.3}=\frac{1}{6}=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4.2.3}=\frac{1}{6} , which is true

For n = k we have

p(k):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)} \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(k+1)(k+2)(k+3)}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}

=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \ \ \ \ \ \ (using \ (i))
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)}{4}+ \frac{1}{k+3} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)^2+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+9+6k)+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+9k+6k^2+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+2k^2+k+8k+4k^2+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{(k+1)^2(k+4)}{4(k+3)} \right )
= \frac{(k+1)((k+1)+3)}{4(k+1+1)(k+1+2)}

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 n

Question:12 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}

Answer:

Let the given statement be p(n) i.e.
p(n):a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}
For n = 1 we have
p(1):a=\frac{a(r^1-1)}{r-1}=\frac{r-1}{r-1}=1 , which is true

For n = k we have

p(k):a+ar+ar^2+...+ar^{k-1}=\frac{a(r^k-1)}{r-1} \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):a+ar+ar^2+...+ar^{k}=a+ar+ar^2+...+ar^{k-1}+ar^{k}

=a.\frac{r^k-1}{r-1}+ar^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{a(r^k-1)+(r-1)ar^{k}}{r-1}
=\frac{ar^k(1+r-1)-a}{r-1}
=\frac{ar^k.r-a}{r-1}
=\frac{a(r^{k+1}-1)}{r-1}

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 n

Question:13 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2

Answer:

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 )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2
For n = 1 we have
p(1):\left ( 1+\frac{3}{1} \right )= 4=(1+1)^2=2^2=4 , which is true

For n = k we have

p(k):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2k+1)}{k^2} \right )=(k+1)^2 \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right ) =\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{2k+1}{k^2} \right )\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right )

=(k+1)^2\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right ) \ \ \ \ \ \ \ \ \ \ (using \ (i))
=(k+1)^2\left ( \frac{{}(k+1)^2+(2(k+1)+1)}{(k+1)^2} \right )
=(k^2+1+2k+2k+2+1)
=(k^2+4k+4)
=(k+2)^2
=(k+1+1)^2

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 n

Question:14 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{n} \right )=(n+1)

Answer:

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 )...\left ( 1+\frac{1}{n} \right )=(n+1)
For n = 1 we have
p(1):\left ( 1+\frac{1}{1} \right )=2=(1+1)=2 , which is true

For n = k we have

p(k):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k} \right )=(k+1) \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k+1} \right ) &nbsnbsp; =\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k} \right )\left ( 1+\frac{1}{k+1} \right )

=(k+1)\left ( 1+\frac{1}{k+1} \right ) \ \ \ \ \ \ \ (using \ (i))
=(k+1)\left ( \frac{k+1+1}{k+1} \right )
=(k+2)
=(k+1+1)
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 n

Question:15 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}

Answer:

Let the given statement be p(n) i.e.
p(n):1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}
For n = 1 we have
p(1):1^2=1=\frac{1(2(1)-1)(2(1)+1)}{3}= \frac{1.1.3}{3}=1 , which is true

For n = k we have

p(k):1^2+3^2+5^2+...+(2k-1)^2=\frac{k(2k-1)(2k+1)}{3} \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1^2+3^2+5^2+...+(2(k+1)-1)^2=1^2+3^2+5^2+...+(2k-1)^2+(2(k+1)-1)^2

=\frac{k(2k-1)(2k+1)}{3}+(2(k+1)-1)^2 \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(2k-1)(2k+1)+3(2(k+1)-1)^2}{3}
=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}
=\frac{(2k+1)(k(2k-1)+3(2k+1))}{3}
=\frac{(2k+1)(2k^2-k+6k+3)}{3}
=\frac{(2k+1)(2k^2+5k+3)}{3}
=\frac{(2k+1)(2k^2+2k+3k+3)}{3}
=\frac{(2k+1)(2k+3)(k+1)}{3}
=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}

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 n

Question:16 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}
For n = 1 we have
p(1):\frac{1}{1.4}=\frac{1}{4}=\frac{1}{(3(1)+1)}=\frac{1}{3+1}=\frac{1}{4} , which is true

For n = k we have

p(k):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{k}{(3k+1)} \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3(k+1)-2)(3(k+1)+1)}=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{(3(k+1)-2)(3(k+1)+1)}

=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)} \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{1}{3k+1}\left ( k+\frac{1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{k(3k+4)+1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{3k^2+4k+1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{3k^2+3k+k+1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{(3k+1)(k+1)}{3k+4} \right )
= \frac{(k+1)}{3k+4}
= \frac{(k+1)}{3(k+1)+1}

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 n

Question:17 Prove the following by using the principle of mathematical induction for all n\in N : \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}
For n = 1 we have
p(1):\frac{1}{3.5}= \frac{1}{15}=\frac{1}{3(2(1)+3)}=\frac{1}{3.5}=\frac{1}{15} , which is true

For n = k we have

p(k):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)} \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2(k+1)+1)(2(k+1)+3)}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{(2(k+1)+1)(2(k+1)+3)}

=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{1}{2k+3}\left ( \frac{k}{3}+\frac{1}{2k+5} \right )
=\frac{1}{2k+3}\left ( \frac{k(2k+5)+3}{3(2k+5)} \right )
=\frac{1}{2k+3}\left ( \frac{2k^2+5k+3}{3(2k+5)} \right )
=\frac{1}{2k+3}\left ( \frac{2k^2+2k+3k+3}{3(2k+5)} \right )
=\frac{1}{2k+3}\left ( \frac{(2k+3)(k+1)}{3(2k+5)} \right )
= \frac{(k+1)}{3(2k+5)}
= \frac{(k+1)}{3(2(k+1)+3)}

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 n

Question:18 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1+2+3+...+n<\frac{1}{8}(2n+1)^2.

Answer:

Let the given statement be p(n) i.e.
p(n):1+2+3+...+n<\frac{1}{8}(2n+1)^2.
For n = 1 we have
p(1):1<\frac{1}{8}(2(1)+1)^2= \frac{1}{8}(3)^2=\frac{9}{8} , which is true

For n = k we have

p(k):1+2+3+...+k<\frac{1}{8}(2k+1)^2 \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1+2+3+...+k+1=1+2+3+...+k+k+1

< \frac{1}{8}\left ( 2k+1 \right )^2+(k+1) \ \ \ \ \ \ \ \ \ (using \ (i))
< \frac{1}{8}\left ( (2k+1)^2+8(k+1) \right )
< \frac{1}{8}\left ( 4k^2+4k+1+8k+8 \right )
< \frac{1}{8}\left ( 4k^2+12k+9\right )
< \frac{1}{8}\left ( 2k+3\right )^2
< \frac{1}{8}\left ( 2(k+1)+1\right )^2

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 n

Question:19 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : n(n+1)(n+5) is a multiple of 3.

Answer:

Let the given statement be p(n) i.e.
p(n):n(n+1)(n+5)
For n = 1 we have
p(1):1(1+1)(1+5)=1.2.6=12 , which is multiple of 3, hence true

For n = k we have

p(k):k(k+1)(k+5) \ \ \ \ \ \ \ -(i) , Let's assume that this is multiple of 3 = 3m
Now,
For n = k + 1 we have
p(k+1):(k+1)((k+1)+1)((k+1)+5)=(k+1)(k+2)((k+5)+1)

=(k+1)(k+2)(k+5)+(k+1)(k+2)
=k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)
=3m+2(k+1)(k+5)+(k+1)(k+2) \ \ \ \ \ \ \ \ (using \ (i))
=3m+(k+1)(2(k+5)+(k+2)) \ \ \ \ \ \ \ \ (using \ (i))
=3m+(k+1)(2k+10+k+2)
=3m+(k+1)(3k+12)
=3m+3(k+1)(k+4)
=3(m+(k+1)(k+4) )
=3l Where \left ( l=(m+(k+1)(k+4) ) \right ) some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is multiple of 3 for all natural numbers n

Question:20 Prove the following by using the principle of mathematical induction for all n\in N : 10^{2n-1}+1 is a divisible by 11.

Answer:

Let the given statement be p(n) i.e.
p(n):10^{2n-1}+1
For n = 1 we have
p(1):10^{2(1)-1}+1= 10^{2-1}+1=10^1+1=11 , which is divisible by 11, hence true

For n = k we have

p(k):10^{2k-1}+1 \ \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by 11 = 11m
Now,
For n = k + 1 we have
p(k+1):10^{2(k+1)-1}+1=10^{2k+2-1}+1
nbsp;
=10^{2k+1}+1
=10^2(10^{2k-1}+1-1)+1
=10^2(10^{2k-1}+1)-10^2+1
=10^2(11m)-100+1 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=100(11m)-99
=11(100m-9)
=11l Where l= (100m-9) some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 11 for all natural numbers n

Question:21 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : x^2^n-y^2^n is divisible by x+y.

Answer:

Let the given statement be p(n) i.e.
p(n):x^2^n-y^2^n
For n = 1 we have
p(1):x^{2(1)}-y^{2(1)}= x^2-y^2=(x-y)(x+y) , which is divisible by (x+y) , hence true (using \ a^2-b^2=(a+b)(a-b))

For n = k we have

p(k):x^{2k}-y^{2k} \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by (x+y) =(x+y)m
Now,
For n = k + 1 we have
p(k+1):x^{2(k+1)}-y^{2(k+1)} =x^{2k}.x^2-y^{2k}.y^2

=x^2(x^{2k}+y^{2k}-y^{2k})-y^{2k}.y^2
=x^2(x^{2k}-y^{2k})+x^2.y^{2k}-y^{2k}.y^2
=x^2(x+y)m+(x^2-y^2)y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=x^2(x+y)m+((x-y)(x+y))y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ a^2-b^2=(a+b)(a-b))
=(x+y)\left ( x^2.m+(x-y).y^{2k} \right )
=(x+y)l where l = (x^2.m+(x-y).y^{2k}) some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by (x+y) for all natural numbers n

Question:22 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 3^{2n+2}-8n-9 is divisible by 8.

Answer:

Let the given statement be p(n) i.e.
p(n):3^{2n+2}-8n-9
For n = 1 we have
p(1):3^{2(1)+2}-8(1)-9= 3^4-8-9=81-17=64=8\times 8 , which is divisible by 8, hence true

For n = k we have

p(k):3^{2k+2}-8k-9 \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by 8 = 8m
Now,
For n = k + 1 we have
p(k+1):3^{2(k+1)+2}-8(k+1)-9=3^{2k+2+2}-8(k+1)-9
=3^{2k+2}.3^2-8k-8-9
=3^2(3^{2k+2}-8k-9+8k+9)-8k-17
=3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17
=9\times 8m+9(8k+9)-8k-17 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=9\times 8m+72k+81-8k-17
=9\times 8m+80k-64
=9\times 8m+8(10k-8)
=8(9m+10k-8)
=8l where l= 9m+10k-8 some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 8 for all natural numbers n

Question:23 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : 41^n-14^n is a multiple of 27.

Answer:

Let the given statement be p(n) i.e.
p(n):41^n-14^n
For n = 1 we have
p(1):41^1-14^1= 41-14= 27 , which is divisible by 27, hence true

For n = k we have

p(k):41^k-14^k \ \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by 27 = 27m
Now,
For n = k + 1 we have
p(k+1):41^{k+1}-14^{k+1}=41^{k}.41-14^{k}.14
=41(41^{k}-14^k+14^k)-14^{k}.14
=41(41^{k}-14^k)+14^k.41-14^{k}.14
=41(27m)+14^k(41-14) \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=41(27m)+14^k.27
=27(41m+14^k)
=27l where l = 41m+14^k some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 27 for all natural numbers n

Question:24 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : (2n+7)<(n+3)^2

Answer:

Let the given statement be p(n) i.e.
p(n):(2n+7)<(n+3)^2
For n = 1 we have
p(1):(2(1)+7)<(1+3)^2\Rightarrow 9< 16 , which is true

For n = k we have

p(k):(2k+7)<(k+3)^2 \ \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is true
Now,
For n = k + 1 we have
p(k+1):(2(k+1)+7)=(2k+2+7)
<(k+3)^2+2 \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
<k^2+9+6k+2
<k^2+6k+11
Now , \ <k^2+6k+11<k^2+8k+16
<(k+4)^2
<((k+1)+3)^2

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 n

More About NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1

Before the introduction of exercise 4.1 Class 11 Maths 8 example problems are given in NCERT chapter 4 of Class 11. Only one exercise is discussed in this chapter, which is Class 11 Maths chapter 4 exercise 4.1. 24 questions are discussed in the Class 11 Maths chapter 4 exercise 4.1. Solving all these questions are important as students may face similar or same questions in the exam paper. Give a good number of try to solve questions before looking to the NCERT solutions for Class 11 Maths chapter 4 exercise 4.1.

Also Read| Principle Of Mathematical Induction Class 11th Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1

  • Solving all the questions of exercise 4.1 Class 11 Maths gives a good idea of the steps involved in solving a given problem using mathematical induction

  • From this 11th class maths exercise 4.1 answers, students can definitely expect a question for the Class 11 final exam. To solve questions in the exam paper it is important to practice NCERT solutions for Class 11 Maths chapter 4 exercise 4.1.

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Key Features of 11th Class Maths Exercise 4.1 Answers

  1. Step-by-step explanations: Detailed ex 4.1 class 11 solutions with a step-by-step breakdown for each problem.

  2. Clarity and precision: Clear and accurate presentation, helping students confidently prepare for exams.

  3. Conceptual understanding: Emphasis on fostering a deep understanding of mathematical concepts.

  4. Curriculum alignment: class 11 maths ex 4.1 solutions closely follow the NCERT curriculum, covering topics as per the official syllabus.

  5. Variety of practice problems: A range of class 11 maths chapter 4 exercise 4.1 exercises provided for students to enhance their problem-solving skills.

  6. Accessibility: Class 11 ex 4.1 solutions are often available for free, making them easily accessible to students.

  7. Format options: PDF versions are typically offered for convenient online and offline access.

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Frequently Asked Question (FAQs)

1. How many topics are given in the NCERT Class 11 chapter 4?

Three topics

2. What are the three topics given in the chapter principle of mathematical induction?

Introduction, motivation and the principle of mathematical induction

3. How many questions are solved in the Class 11 Maths chapter 4 exercise 4.1?

24 questions

4. What number of solved examples are given in the NCERT Class 11 Maths Chapter 4 - Principle of Mathematical Induction?

Eight examples are solved before the Exercise 4.1 Class 11 Maths

5. What is the relation between the principle of mathematical induction and Peano’s axioms?

It is a restatement of one of Peano’s axioms

6. Who was the first person to name and define mathematical induction?

De Morgan

7. Which mathematician is credited with the origin of the principle of mathematical induction?

Blaise Pascal

8. Why do we solve NCERT Solutions for Class 11 Maths chapter 4 exercise 4.1?

The basic aim of solving the NCERT exercise is to understand the concepts well and to get an insight in to different types of questions that uses the concepts studied. Other than this Class 11th Maths chapter 4 exercise 4.1 will help in exams also. 

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