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

Updated on Apr 13, 2020 - 5:05 p.m. IST by Ravindra Pindel

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction : In this chapter, you will learn a mathematical technique that can be used to prove a theorem, formula or statement true for every natural number. It is called mathematical induction. In this article, you will get NCERT solutions for class 11 maths chapter 4 principle of mathematical induction. This chapter starts with an introduction to deductive and inductive reasoning.

Deductive reasoning uses some statement of facts and deducts another fact.

For example, consider the following statements a and b are true

a) All men are mortal

b) Raju is a man

From the above two statements we can deduct the fact c) Raju is mortal that is using a and b, c is established.

In inductive reasoning, we look for a pattern and obtaining a conclusion. That is induction means generalization from particular cases. For example, if we have data on the population of a city for the last 10 years. Based on the trend of population growth in the last 10 years we can predict the population of the next coming 2 years. This problem comes under induction. In CBSE NCERT solutions for class 11 maths chapter 4 principle of mathematical induction, you will learn a two-step process to prove any formula, statement or theorem. There are 24 problems in this chapter. All these problems are explained in solutions of NCERT for class 11 maths chapter 4 principle of mathematical induction in a step-by-step manner. It will be very easy for you to understand the concepts. Check all NCERT solutions from class 6 to 12 to learn CBSE science and maths.

The main topics of NCERT Class 11 Maths Chapter 4 Principle of Mathematical Induction are

4.1 Introduction

4.2 Motivation

4.3 The Principle of Mathematical Induction

NCERT solutions for class 11 maths chapter 4 principle of mathematical induction-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)2^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 have
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

NCERT solutions for class 11 mathematics

chapter-1

NCERT solutions for class 11 maths chapter 1 Sets

chapter-2

Solutions of NCERT for class 11 chapter 2 Relations and Functions

chapter-3

CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions

chapter-4

NCERT solutions for class 11 Maths chapter 4 Principle of Mathematical Induction

chapter-5

Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations

chapter-6

CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities

chapter-7

NCERT solutions for class 11 maths chapter 7 Permutation and Combinations

chapter-8

Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem

chapter-9

CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series

chapter-10

NCERT solutions for class 11 maths chapter 10 Straight Lines

chapter-11

Solutions of NCERT for class 11 maths chapter 11 Conic Section

chapter-12

CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry

chapter-13

NCERT solutions for class 11 maths chapter 13 Limits and Derivatives

chapter-14

Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning

chapter-15

CBSE NCERT solutions for class 11 maths chapter 15 Statistics

chapter-16

NCERT solutions for class 11 maths chapter 16 Probability

NCERT solutions for class 11- Subject wise

Solutions of NCERT for class 11 biology

CBSE NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

Solutions of NCERT for Class 11 physics

All the above questions explained in NCERT solutions for class 11 maths chapter 4 principle of mathematical induction followed the same common process stated as given below.

Suppose there is a  statement q(n) involving the natural number n

such that (i) The statement is true for n = 1, that is, q(1) is true

(ii) If the statement is true for n = k (k is some positive integer), then the statement is also true for n = k + 1

Note:  Property (ii)  does not assert that the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1.

There are 24 problems solved in this article. First, try to solve by yourself. If you are finding difficulties, you can take help from CBSE NCERT solutions for class 11 maths chapter 4 principle of mathematical induction.

Happy learning !!!

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