***According to the Revised CBSE Syllabus 2020-21, this chapter has been removed.**

NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction are given in an understandable way by the faculty at Goyanka Maths Study. Students learn about the Principle of Mathematical Induction and its application in detail through this chapter. By practising all the problems present in the NCERT Solutions, students can easily score maximum marks in the examinations.

Principle of Mathematical Induction is a specific technique used to prove certain mathematically accepted statements in algebra and in other applications of Mathematics, such as inductive and deductive reasoning. NCERT Solutions of Goyanka Maths Study cover all these concepts and help in scoring full marks in this chapter. These solutions are useful for further studies and for those who are preparing for competitive exams. NCERT Solutions For Class 11 Maths are very accurate and make it easy for the students to crack the exam with good marks.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction

### Access Answers of Maths NCERT Class 11 Chapter 4- Principle of Mathematical Induction

### Access NCERT Solutions for Class 11 Maths Chapter 4

Exercise 4.1 page: 94

**Prove the following by using the principle of mathematical induction for all n ∈ N:**

**1.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

**2.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

**3.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

**4.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**5.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**6.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**7.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**8. 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2**

**Solution:**

We can write the given statement as

P (*n*): 1.2 + 2.22 + 3.22 + … + *n*.2*n* = (*n* – 1) 2*n*+1 + 2

If n = 1 we get

P (1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2

Which is true.

Consider P (k) be true for some positive integer k

1.2 + 2.22 + 3.22 + … + *k.*2*k* = (*k* – 1) 2*k* + 1 + 2 … (i)

Now let us prove that P (k + 1) is true.

Here

P (k + 1) is true whenever P (k) is true.

**9.**

**Solution:**

We can write the given statement as

P (k + 1) is true whenever P (k) is true.

**10.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**11.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**12.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**13.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**14.**

**Solution:**

By further simplification

= (k + 1) + 1

P (k + 1) is true whenever P (k) is true.

**15.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**16.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**17.**

**Solution:**

P (k + 1) is true whenever P (k) is true.

**18.**

**Solution:**

We can write the given statement as

P (k + 1) is true whenever P (k) is true.

**19. n (n + 1) (n + 5) is a multiple of 3**

**Solution:**

We can write the given statement as

P (n): n (n + 1) (n + 5), which is a multiple of 3

If n = 1 we get

1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

Which is true.

Consider P (k) be true for some positive integer k

k (k + 1) (k + 5) is a multiple of 3

k (k + 1) (k + 5) = 3m, where m ∈ N …… (1)

Now let us prove that P (k + 1) is true.

Here

(k + 1) {(k + 1) + 1} {(k + 1) + 5}

We can write it as

= (k + 1) (k + 2) {(k + 5) + 1}

By multiplying the terms

= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)

So we get

= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)

Substituting equation (1)

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

By multiplication

= 3m + (k + 1) {2k + 10 + k + 2}

On further calculation

= 3m + (k + 1) (3k + 12)

Taking 3 as common

= 3m + 3 (k + 1) (k + 4)

We get

= 3 {m + (k + 1) (k + 4)}

= 3 × q where q = {m + (k + 1) (k + 4)} is some natural number

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

P (k + 1) is true whenever P (k) is true.

**20. 102 n – 1 + 1 is divisible by 11**

**Solution:**

We can write the given statement as

P (*n*): 102*n* – 1 + 1 is divisible by 11

If n = 1 we get

P (1) = 102.1 – 1 + 1 = 11, which is divisible by 11

Which is true.

Consider P (k) be true for some positive integer k

102*k* – 1 + 1 is divisible by 11

102*k* – 1 + 1 = 11*m*, where *m* ∈ N** **…… (1)

Now let us prove that P (k + 1) is true.

Here

10 2 (k + 1) – 1 + 1

We can write it as

= 10 2k + 2 – 1 + 1

= 10 2k + 1 + 1

By addition and subtraction of 1

= 10 2 (102k-1 + 1 – 1) + 1

We get

= 10 2 (102k-1 + 1) – 102 + 1

Using equation 1 we get

= 102. 11m – 100 + 1

= 100 × 11m – 99

Taking out the common terms

= 11 (100m – 9)

= 11 r, where r = (100m – 9) is some natural number

10 2(k + 1) – 1 + 1 is divisible by 11

P (k + 1) is true whenever P (k) is true.

**21. x2n – y2n is divisible by x + y**

**Solution:**

We can write the given statement as

P (*n*): *x*2*n* – *y*2*n* is divisible by* x *+ *y*

If n = 1 we get

P (1) = *x*2 × 1 – *y*2 × 1 = *x*2 – *y*2 = (*x *+ *y*) (*x* – *y*), which is divisible by (x + y)

Which is true.

Consider P (k) be true for some positive integer k

*x*2*k* – *y*2*k* is divisible by* x *+ *y*

*x*2*k* – *y*2*k* = *m* (*x *+ *y*), where *m* ∈ N** **…… (1)

Now let us prove that P (k + 1) is true.

Here

x 2(k + 1) – y 2(k + 1)

We can write it as

= x 2k . x2 – y2k . y2

By adding and subtracting y2k we get

= x2 (x2k – y2k + y2k) – y2k. y2

From equation (1) we get

= x2 {m (x + y) + y2k} – y2k. y2

By multiplying the terms

= m (x + y) x2 + y2k. x2 – y2k. y2

Taking out the common terms

= m (x + y) x2 + y2k (x2 – y2)

Expanding using formula

= m (x + y) x2 + y2k (x + y) (x – y)

So we get

= (x + y) {mx2 + y2k (x – y)}, which is a factor of (x + y)

P (k + 1) is true whenever P (k) is true.

**22. 32 n + 2 – 8n – 9 is divisible by 8**

**Solution:**

We can write the given statement as

P (*n*): 32*n* + 2 – 8*n* – 9 is divisible by 8

If n = 1 we get

P (1) = 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8

Which is true.

Consider P (k) be true for some positive integer k

32*k* + 2 – 8*k* – 9 is divisible by 8

32*k* + 2 – 8*k* – 9 = 8*m*, where *m* ∈ N** **…… (1)

Now let us prove that P (k + 1) is true.

Here

3 2(k + 1) + 2 – 8 (k + 1) – 9

We can write it as

= 3 2k + 2 . 32 – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 32 (32k + 2 – 8k – 9 + 8k + 9) – 8k – 17

On further simplification

= 32 (32k + 2 – 8k – 9) + 32 (8k + 9) – 8k – 17

From equation (1) we get

= 9. 8m + 9 (8k + 9) – 8k – 17

By multiplying the terms

= 9. 8m + 72k + 81 – 8k – 17

So we get

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

So 3 2(k + 1) + 2 – 8 (k + 1) – 9 is divisible by 8

P (k + 1) is true whenever P (k) is true.

**23. 41 n – 14n is a multiple of 27**

**Solution:**

We can write the given statement as

P (*n*):41*n* – 14*n*is a multiple of 27

If n = 1 we get

P (1) = 411 – 141 = 27, which is a multiple by 27

Which is true.

Consider P (k) be true for some positive integer k

41*k* – 14*k*is a multiple of 27

41*k* – 14*k* = 27*m*, where *m* ∈ N** **…… (1)

Now let us prove that P (k + 1) is true.

Here

41k + 1 – 14 k + 1

We can write it as

= 41k. 41 – 14k. 14

By adding and subtracting 14k we get

= 41 (41k – 14k + 14k) – 14k. 14

On further simplification

= 41 (41k – 14k) + 41. 14k – 14k. 14

From equation (1) we get

= 41. 27m + 14k ( 41 – 14)

By multiplying the terms

= 41. 27m + 27. 14k

By taking out the common terms

= 27 (41m – 14k)

= 27r, where r = (41m – 14k) is a natural number

So 41k + 1 – 14k + 1 is a multiple of 27

P (k + 1) is true whenever P (k) is true.

**24. (2 n +7) < (n + 3)2**

**Solution:**

We can write the given statement as

P(*n*): (2*n *+7) < (*n* + 3)2

If n = 1 we get

2.1 + 7 = 9 < (1 + 3)2 = 16

Which is true.

Consider P (k) be true for some positive integer k

(2*k* + 7) < (*k* + 3)2 … (1)

Now let us prove that P (k + 1) is true.

Here

{2 (k + 1) + 7} = (2k + 7) + 2

We can write it as

= {2 (k + 1) + 7}

From equation (1) we get

(2k + 7) + 2 < (k + 3)2 + 2

By expanding the terms

2 (k + 1) + 7 < k2 + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k2 + 6k + 11

Here k2 + 6k + 11 < k2 + 8k + 16

We can write it as

2 (k + 1) + 7 < (k + 4)2

2 (k + 1) + 7 < {(k + 1) + 3}2

P (k + 1) is true whenever P (k) is true.