NCERT Exemplar Class 11 Maths Solutions for Chapter 4 - Principle Of Mathematical Induction

NCERT Exemplar Class 11 Maths Solutions for Chapter 4 - PDF Download Free Download

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction, are provided here for students to practice and prepare for the exam. For students to make it easy for them to learn the concepts explained in this chapter, NCERT exemplar problems with solutions for chapter 4 Principle of Mathematical Induction are provided here. These exemplar problems are prepared by our experts in accordance with CBSE syllabus (2018-2019) and curriculum.

Class 11 Maths NCERT Exemplar Problems on Principle of Mathematical Induction

In chapter 4, students will learn and solve exemplar problems based on the Principle of mathematical induction.  If S(n) is a given statement, where n is the natural numbers, then;

  • The statement is true for n = 1, i.e., S(1) is true
  • If the statement is true for n = k (where i is some positive integer), then the statement is also true for n = i + 1, i.e., the truth of S(k) implies the truth of S(i + 1).

Then, S(n) is true for all natural numbers n.

The Principle of Mathematical Induction is a technique used as a proof in mathematics. The major function of this principle is used as a defining property for every natural number such as 1,2,3,4 etc. The principle can effectively be proved with the help of two cases which are the base case and the induction step. It is defined as the property which holds for the number 0. On the other hand, the property for one natural number ‘n’ and the next natural number is ‘n+1’.

Download PDF Of NCERT Exemplar Class 10 Maths Chapter 4 – Principle Of Mathematical Induction

 

ncert exemplar solutions for class 11 maths chapter 4
ncert exemplar solutions for class 11 maths chapter 4
ncert exemplar solutions for class 11 maths chapter 4
ncert exemplar solutions for class 11 maths chapter 4
ncert exemplar solutions for class 11 maths chapter 4
ncert exemplar solutions for class 11 maths chapter 4
ncert exemplar solutions for class 11 maths chapter 4

 

Access Answers of NCERT Exemplar Class 10 Maths Chapter 4 – Principle Of Mathematical Induction

Exercise Page No: 70

Short Answer Type

1. Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.

Solution:

According to the question,

P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true

Let P(n) be 2n < n!

So, the examples of the given statements are,

P(0) ⇒ 20 < 0!

i.e 1 < 1 ⇒ not true

P(1) ⇒ 21 < 1!

i.e 2 < 1 ⇒ not true

P(2) ⇒ 22 < 2!

i.e 4 < 2 ⇒ not true

P(3) ⇒ 23 < 3!

i.e 8 < 6 ⇒ not true

P(4) ⇒ 24 < 4!

i.e 16 < 24 ⇒ true

P(5) ⇒ 25 < 5!

i.e 32 < 60 ⇒ true, etc.

2. Give an example of a statement P(n) which is true for all n. Justify your answer.

Solution:

According to the question,

P(n) which is true for all n.

Let P(n) be,

NCERT Exemplar Solutions Class 11 Maths Chapter 4-1

⇒ P(k) is true for all k.

Therefore, P(n) is true for all n.

Prove each of the statements in Exercises 3 to 16 by the Principle of Mathematical Induction:

3. 4n – 1 is divisible by 3, for each natural number n.

Solution:

According to the question,

P(n) = 4n – 1 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 40 – 1 = 0 which is divisible by 3.

P(1) = 41 – 1 = 3 which is divisible by 3.

P(2) = 42 – 1 = 15 which is divisible by 3.

P(3) = 43 – 1 = 63 which is divisible by 3.

Let P(k) = 4k – 1 be divisible by 3,

So, we get,

⇒ 4k – 1 = 3x.

Now, we also get that,

⇒  P(k+1) = 4k+1 – 1

= 4(3x + 1) – 1

= 12x + 3 is divisible by 3.

⇒ P(k+1) is true when P(k) is true

Therefore, by Mathematical Induction,

P(n) = 4n – 1 is divisible by 3 is true for each natural number n.

4. 23n – 1 is divisible by7, for all natural numbers n.

Solution:

According to the question,

P(n) = 23n – 1 is divisible by 7.

So, substituting different values for n, we get,

P(0) = 20 – 1 = 0 which is divisible by 7.

P(1) = 23 – 1 = 7 which is divisible by 7.

P(2) = 26 – 1 = 63 which is divisible by 7.

P(3) = 29 – 1 = 512 which is divisible by 7.

Let P(k) = 23k – 1 be divisible by 7

So, we get,

⇒ 23k – 1 = 7x.

Now, we also get that,

⇒  P(k+1) = 23(k+1) – 1

= 23(7x + 1) – 1

= 56x + 7

= 7(8x + 1) is divisible by 7.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 23n – 1 is divisible by7, for all natural numbers n.

5. n3 – 7n + 3 is divisible by 3, for all natural numbers n.

Solution:

According to the question,

P(n) = n3 – 7n + 3 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 03 – 7×0 + 3 = 3 which is divisible by 3.

P(1) = 13 – 7×1 + 3 = −3 which is divisible by 3.

P(2) = 23 – 7×2 + 3 = −3 which is divisible by 3.

P(3) = 33 – 7×3 + 3 = 9 which is divisible by 3.

Let P(k) = k3 – 7k + 3 be divisible by 3

So, we get,

⇒ k3 – 7k + 3 = 3x.

Now, we also get that,

⇒  P(k+1) = (k+1)3 – 7(k+1) + 3

= k3 + 3k2 + 3k + 1 – 7k – 7 + 3

= 3x + 3(k2 + k – 2) is divisible by 3.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n3 – 7n + 3 is divisible by 3, for all natural numbers n.

6. 32n – 1 is divisible by 8, for all natural numbers n.

Solution:

According to the question,

P(n) = 32n – 1 is divisible by 8.

So, substituting different values for n, we get,

P(0) = 30 – 1 = 0 which is divisible by 8.

P(1) = 32 – 1 = 8 which is divisible by 8.

P(2) = 34 – 1 = 80 which is divisible by 8.

P(3) = 36 – 1 = 728 which is divisible by 8.

Let P(k) = 32k – 1 be divisible by 8

So, we get,

⇒ 32k – 1 = 8x.

Now, we also get that,

⇒  P(k+1) = 32(k+1) – 1

= 32(8x + 1) – 1

= 72x + 8 is divisible by 8.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 32n – 1 is divisible by 8, for all natural numbers n.

7. For any natural number n, 7n – 2n is divisible by 5.

Solution:

According to the question,

P(n) = 7n – 2n is divisible by 5.

So, substituting different values for n, we get,

P(0) = 70 – 20 = 0 Which is divisible by 5.

P(1) = 71 – 21 = 5 Which is divisible by 5.

P(2) = 72 – 22 = 45 Which is divisible by 5.

P(3) = 73 – 23 = 335 Which is divisible by 5.

Let P(k) = 7k – 2k be divisible by 5

So, we get,

⇒ 7k – 2k = 5x.

Now, we also get that,

⇒  P(k+1)= 7k+1 – 2k+1

= (5 + 2)7k – 2(2k)

= 5(7k) + 2 (7k – 2k)

= 5(7k) + 2 (5x) Which is divisible by 5.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 7n – 2n is divisible by 5 is true for each natural number n.

8. For any natural number n, xn – yn is divisible by x – y, where x integers with x ≠ y.

Solution:

According to the question,

P(n) = xn – yn is divisible by x – y, x integers with x ≠ y.

So, substituting different values for n, we get,

P(0) = x0 – y0 = 0 Which is divisible by x − y.

P(1) = x − y Which is divisible by x − y.

P(2) = x2 – y2

= (x +y)(x−y) Which is divisible by x−y.

P(3) = x3 – y3

= (x−y)(x2+xy+y2) Which is divisible by x−y.

Let P(k) = xk – yk be divisible by x – y;

So, we get,

⇒ xk – yk = a(x−y).

Now, we also get that,

⇒  P(k+1) = xk+1 – yk+1

= xk(x−y) + y(xk−yk)

= xk(x−y) +y a(x−y) Which is divisible by x − y.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) xn – yn is divisible by x – y, where x integers with x ≠ y which is true for any natural number n.

9. n3 – n is divisible by 6, for each natural number n.

Solution:

According to the question,

P(n) = n3 – n is divisible by 6.

So, substituting different values for n, we get,

P(0) = 03 – 0 = 0 Which is divisible by 6.

P(1) = 13 – 1 = 0 Which is divisible by 6.

P(2) = 23 – 2 = 6 Which is divisible by 6.

P(3) = 33 – 3 = 24 Which is divisible by 6.

Let P(k) = k3 – k be divisible by 6.

So, we get,

⇒ k3 – k = 6x.

Now, we also get that,

⇒  P(k+1) = (k+1)3 – (k+1)

= (k+1)(k2+2k+1−1)

= k3 + 3k2 + 2k

= 6x+3k(k+1) [n(n+1) is always even and divisible by 2]

= 6x + 3×2y Which is divisible by 6.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n3 – n is divisible by 6, for each natural number n.

10. n(n2 + 5) is divisible by 6, for each natural number n.

Solution:

According to the question,

P(n) = n(n2 + 5) is divisible by 6.

So, substituting different values for n, we get,

P(0) = 0(02 + 5) = 0 Which is divisible by 6.

P(1) = 1(12 + 5) = 6 Which is divisible by 6.

P(2) = 2(22 + 5) = 18 Which is divisible by 6.

P(3) = 3(32 + 5) = 42 Which is divisible by 6.

Let P(k) = k(k2 + 5) be divisible by 6.

So, we get,

⇒ k(k2 + 5) = 6x.

Now, we also get that,

⇒  P(k+1) = (k+1)((k+1)2 + 5) = (k+1)(k2+2k+6)

= k3 + 3k2 + 8k + 6

= 6x+3k2+3k+6

= 6x+3k(k+1)+6[n(n+1) is always even and divisible by 2]

= 6x + 3×2y + 6 Which is divisible by 6.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n(n2 + 5) is divisible by 6, for each natural number n.

11. n2 < 2n for all natural numbers n ≥ 5.

Solution:

According to the question,

P(n) is n2 < 2n  for n≥5

Let P(k) = k2 < 2k be true;

⇒ P(k+1) = (k+1)2

= k2 + 2k + 1

2k+1 = 2(2k) > 2k2

Since, n2 > 2n + 1 for n ≥3

We get that,

 k2 + 2k + 1 < 2k2

⇒ (k+1)2 < 2(k+1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n2 < 2n is true for all natural numbers n ≥ 5.

12. 2n < (n + 2)! for all natural number n.

Solution:

According to the question,

P(n) is 2n < (n + 2)!

So, substituting different values for n, we get,

P(0) ⇒ 0 < 2!

P(1) ⇒ 2 < 3!

P(2) ⇒ 4 < 4!

P(3) ⇒ 6 < 5!

Let P(k) = 2k < (k + 2)! is true;

Now, we get that,

⇒ P(k+1) = 2(k+1) ((k+1)+2))!

We know that,

[(k+1)+2)! = (k+3)! = (k+3)(k+2)(k+1)……………3×2×1]

But, we also know that,

= 2(k+1) × (k+3)(k+2)……………3×1 > 2(k+1)

Therefore, 2(k+1) < ((k+1) + 2)!

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 2n < (n + 2)! Is true for all natural number n.

13. √n < 1/√1 + 1/√2 + … 1/√n, for all natural numbers n ≥ 2.

Solution:

According to the question,

NCERT Exemplar Solutions Class 11 Maths Chapter 4-2

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

√n < 1/√1 + 1/√2 + … 1/√n, for all natural numbers n ≥ 2

14. 2 + 4 + 6 + …+ 2n = n2 + n for all natural numbers n.

Solution:

According to the question,

P(n) is 2 + 4 + 6 + …+ 2n = n2 + n.

So, substituting different values for n, we get,

P(0) = 0 = 02 + 0 Which is true.

P(1) = 2 = 12 + 1 Which is true.

P(2) = 2 + 4 = 22 + 2 Which is true.

P(3) = 2 + 4 + 6 = 32 + 2 Which is true.

Let P(k) = 2 + 4 + 6 + …+ 2k = k2 + k be true;

So, we get,

⇒ P(k+1) is 2 + 4 + 6 + …+ 2k + 2(k+1) = k2 + k + 2k +2

= (k2 + 2k +1) + (k+1)

= (k + 1)2 + (k + 1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

2 + 4 + 6 + …+ 2n = n2 + n is true for all natural numbers n.

15. 1 + 2 + 22 + … 2n = 2n+1 – 1 for all natural numbers n.

Solution:

According to the question,

P(n) is 1 + 2 + 22 + … 2n = 2n+1 – 1.

So, substituting different values for n, we get,

P(0) = 1 = 20+1 − 1 Which is true.

P(1) = 1 + 2 = 3 = 21+1 − 1 Which is true.

P(2) = 1 + 2 + 22 = 7 = 22+1 − 1 Which is true.

P(3) = 1 + 2 + 22 + 23 = 15 = 23+1 − 1 Which is true.

Let P(k) = 1 + 2 + 22 + … 2k = 2k+1 – 1 be true;

So, we get,

⇒ P(k+1) is 1 + 2 + 22 + … 2k + 2k+1 = 2k+1 – 1 + 2k+1

= 2×2k+1 – 1

= 2(k+1)+1 – 1

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

1 + 2 + 22 + … 2n = 2n+1 – 1 is true for all natural numbers n.


The solved questions in exemplar of Maths class 11 chapter 4 will help students to know the methods for solving problems based on the Principle of Mathematical Induction. We are also providing on our website, withNCERT solutions, notes and question papers online for free for 11th standard students. Solving previous year question papers and sample papers of 11th class Maths subject, will let them know the question pattern and marking scheme for chapter 4, Principle of Mathematical Induction.

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