NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4

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.

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

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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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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
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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

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

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

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

Key Features of 11th Class Maths Exercise 4.1 Answers

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Clarity and precision: Clear and accurate presentation, helping students confidently prepare for exams.
Conceptual understanding: Emphasis on fostering a deep understanding of mathematical concepts.
Curriculum alignment: class 11 maths ex 4.1 solutions closely follow the NCERT curriculum, covering topics as per the official syllabus.
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NCERT Solutions of Class 11 Subject Wise

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Frequently Asked Questions (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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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

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

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

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