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

Q: What are the three topics given in the chapter principle of mathematical induction?

Introduction, motivation and the principle of mathematical induction

Q: 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

Q: What is the relation between the principle of mathematical induction and Peano’s axioms?

It is a restatement of one of Peano’s axioms

Q: 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.

Edited By Safeer PP | Updated on Jul 6, 2022 - 5:29 p.m. IST

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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Principle Of Mathematical Induction Class 11 Chapter4 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 have $p(k):1.2+2.2^2+3.2^3+...+k.2^k=(k-1)2^{k^+1}+2 \ \ \ \ \ \ -(i)$ $p(1):1.2=2=(1-1)2^{1^+1}+2 = 2$ , 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

NCERT Solutions of Class 11 Subject Wise

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Frequently Asked Question (FAQs) - NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

Question: Why do we solve NCERT Solutions for Class 11 Maths chapter 4 exercise 4.1?

Answer:

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.

Question: Which mathematician is credited with the origin of the principle of mathematical induction?

Answer:

Blaise Pascal

Question: Who was the first person to name and define mathematical induction?

Answer:

De Morgan

Question: What is the relation between the principle of mathematical induction and Peano’s axioms?

Answer:

It is a restatement of one of Peano’s axioms

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

Answer:

Eight examples are solved before the Exercise 4.1 Class 11 Maths

Question: How many questions are solved in the Class 11 Maths chapter 4 exercise 4.1?

Answer:

24 questions

Question: What are the three topics given in the chapter principle of mathematical induction?

Answer:

Introduction, motivation and the principle of mathematical induction

Question: How many topics are given in the NCERT Class 11 chapter 4?

Answer:

Three topics

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

Principle Of Mathematical Induction Class 11 Chapter4 Exercise 4.1

Question:1 Prove the following by using the principle of mathematical induction for all :

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Frequently Asked Question (FAQs) - NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

Question: Why do we solve NCERT Solutions for Class 11 Maths chapter 4 exercise 4.1?

Question: Which mathematician is credited with the origin of the principle of mathematical induction?

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

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Question: What are the three topics given in the chapter principle of mathematical induction?

Question: How many topics are given in the NCERT Class 11 chapter 4?

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