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

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

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

?Chapter – 4?

✍ Principle of Mathematical Induction✍

?Exercise 4.1?

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

5.

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.

6.

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.

7.

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.

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

9.

Solution: We can write the given statement as

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.

10.

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.

11.

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.

12.

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.

13.

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.

14.

Solution:

By further simplification
= (k + 1) + 1
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.

15.

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.

16.

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.

17.

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.

18.

Solution: We can write the given statement as

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.

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) = 3 m, 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)
= 3 m + (k + 1) {2 (k + 5) + (k + 2)}
By multiplication
= 3 m + (k + 1) {2 k + 10 + k + 2}
On further calculation
= 3 m + (k + 1) (3 k + 12)
Taking 3 as common
= 3 m + 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.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

20. 10 ^{2n – 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) = 10^{2.1 – 1} + 1 = 11, which is divisible by 11
Which is true.
Consider P (k) be true for some positive integer k
10 ^{2k – 1} + 1 is divisible by 11
10 ^{2k – 1} + 1 = 11m, 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 (10 ^2k-1 + 1 – 1) + 1
We get
= 10² (10 ^2k-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.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

21. x2n – y2n is divisible by x + y

Solution: We can write the given statement as
P (n): x²ⁿ – y ²ⁿ is divisible by x + y
If n = 1 we get
P (1) = x^2 × 1 – y^2 × 1 = x² – y² = (x + y) (x – y), which is divisible by (x + y)
Which is true.
Consider P (k) be true for some positive integer k
x^2k – y^2k is divisible by x + y
x^2k – y^2k = 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 – y^2k . y²
By adding and subtracting y2k we get
= x² (x^2k – y^2k + y^2k) – y^2k. y²
From equation (1) we get
= x² {m (x + y) + y^2k} – y^2k. y²
By multiplying the terms
= m (x + y) x² + y^2k. x² – y^2k. y²
Taking out the common terms
= m (x + y) ^x2 + y^2k (x² – y²)
Expanding using formula
= m (x + y) x² + y^2k (x + y) (x – y)
So we get
= (x + y) {m^x2 + y^2k (x – y)}, which is a factor of (x + y)
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.

22. 3^{2n + 2} – 8n – 9 is divisible by 8

Solution: We can write the given statement as
P (n): 3^{2n + 2} – 8n – 9 is divisible by 8
If n = 1 we get
P (1) = 33^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8
Which is true.
Consider P (k) be true for some positive integer k
3^{2k + 2} – 8k – 9 is divisible by 8
3^{2k + 2} – 8k – 9 = 8m, 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 (3^{2k + 2} – 8k – 9 + 8k + 9) – 8k – 17
On further simplification
= 3² (3^{2k + 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.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

23. 41n – 14n is a multiple of 27

Solution: We can write the given statement as
P (n):41ⁿ – 14ⁿis 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 = 27m, where m ∈ N …… (1)
Now let us prove that P (k + 1) is true.
Here
41^{k + 1} – 14 ^{k + 1}
We can write it as
= 41^k. 41 – 14^k. 14
By adding and subtracting 14^k we get
= 41 (41^k – 14^k + 14^k) – 14^k. 14
On further simplification
= 41 (41^k – 14^k) + 41. 14^k – 14^k. 14
From equation (1) we get
= 41. 27m + 14^k ( 41 – 14)
By multiplying the terms
= 41. 27m + 27. 14^k
By taking out the common terms
= 27 (41m – 14^k)
= 27r, where r = (41m – 14^k) is a natural number
So 41^{k + 1} – 14^{k + 1} is a multiple of 27
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.

24. (2n +7) < (n + 3)²

Solution: We can write the given statement as
P(n): (2n +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
(2k + 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.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.