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Solutions ⇒ Class 11th ⇒ Mathematics ⇒ 4. Principle Of Mathematical Induction

Solutions 4. Principle Of Mathematical Induction - Exercise 4.1 | Class 11 Mathematics - Toppers Study

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4. Principle Of Mathematical Induction

| Exercise 4.1 |

Solutions 4. Principle Of Mathematical Induction - Exercise 4.1 | Class 11 Mathematics - Toppers Study


Chapter 4. Principle of Mathematical induction

Exercise 4.1

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

Solution:

Let the given statement be P(n), i.e.,

LHS = RHS

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n. 

Solution:

Let the given statement be P(n), i.e.,

LHS = RHS

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n. 

 

Solution: Let the given statement be P(n), so

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Q19. n (n + 1) (n + 5) is a multiple of 3.

Solution:

Let the given statement be P(n), so

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

For n = 1, so we have;

n (n + 1) (n + 5) = 1 × 2 × 6 = 12 = 3 × 4

P(n) is true for n = 1

Assume that P(k) is also true for some positive integer k.

k(k + 1) (k + 5)

= k3 + 6k2 + 5 k = 3m (say)   ……………….. (1)

Now, we shall prove that P(k + 1) is true whenever P(k) is true

Replacing k by k + 1

    k + 1 (k + 2) (k + 6)

= (k + 1) (k2 + 8k + 12)

= k (k2 + 8k + 12) + 1(k2 + 8k + 12)

= k3 + 8k2 + 12k + k2 + 8k + 12

= k3 + 9k2 + 20k + 12

=( k3 + 6k2 + 5 k) + 3k2 + 15k + 12

= 3m + 3k2 + 15k + 12     from (1)

= 3(m + k2 + 5k + 4)     

∴  k + 1 (k + 2) (k + 6) is multiple of 3

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.

Q20.  102n - 1  + 1 is divisible by 11.

Solution:

Let the given statement be P(n), so

P(n) : 102n - 1  + 1 is divisible by 11.

For n = 1, so we have;

102n - 1  + 1 = 102×1 - 1 + 1 = 10 + 1 = 11

P(n) is true for n = 1

Assume that P(k) is also true for some positive integer k.

102k- 1  + 1  = 11m say

102k- 1  = 11m - 1      ……………… (1)

We shall prove that P(k + 1) is true whenever P(k) is true

replacing k by k + 1 we have

     102k - 1  + 1

=  102k + 1  + 1

= 102k × 101 + 1 

= {102k - 1 × 100 + 1}  

= {(11m - 1)× 100 + 1}     from equation (1)

= 1100m - 100+ 1   

= 1100m - 99

= 11(100m - 9)     

102n - 1  + 1 is divisible by 11

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.

Q21.  x2n y2n is divisible by x + y

Solution: Let the given statement be P(n), so

P(n) : x2n – y2n is divisible by x + y

Putting n = 1 we have,

x2n – y2n = x2 - y2 = (x + y) (x - y)

P(n) is true for n = 1

Assume that P(k) is also true for some positive integer k or

x2k – y2k is divisible by (x + y)

So, x2k – y2k = m( x + y)

Or  x2k = m( x + y) + y2k     …………. (1)

We shall prove that P(k + 1) is true whenever P(k) is true

replacing k by k + 1 we have

    x2k + 2 – y2k + 2

= x2k . x2  – y2k .y2

Putting the value of x2k from (1)

= {m( x + y) + y2k} x2  – y2k .y2

= m( x + y) x2 + y2k. x2  – y2k .y2

= m( x + y) x2 + y2k (x2  – y2)

= m( x + y) x2 + y2k (x + y) ( x - y)

= ( x + y) [mx2 + y2k ( x - y)]

 x2n – y2n is divisible by x + y

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.

Q22.  32n+2 – 8n – 9 is divisible by 8

Solution: Let the given statement be P(n), so

P(n) : 32n+2 – 8n – 9 is divisible by 8
Putting n =1

P(1) : 32×1+2 – 8 × 1 – 9 = 81 - 17 = 64 = 8 × 8

Which is divisible by 8

P(1) is true

Assume that P(k) is also true for some positive integer k

 32k + 2 – 8k – 9

 32k + 2 – 8k – 9 is divisible by 8
 32k + 2 – 8k – 9 = 8m

Or 32k + 2 = 8m + 8k + 9     ……………. (1)

We shall prove that P(k + 1) is true whenever P(k) is true

replacing k by k + 1 we have

    32k + 4 – 8k 8  – 9

= 32k + 4 – 8k 17

= 32k + 2 × 32 – 8k 17

= (8m + 8k + 9)× 9 – 8k 17

= 72m + 72k + 81 – 8k 17

= 72m + 64k + 64

= 8(9m + 8k + 8)

∴  32n+2 – 8n – 9 is divisible by 8

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.

Q23. 41n – 14n is a multiple of 27.

Solution: Let the given statement be P(n), so

P(n) : 41n – 14n is a multiple of 27
Putting n = 1

P(1): 41n – 14n = 41 – 14 = 27

P(1) is true

Assume that P(k) is also true for some positive integer k

41k – 14k = 27

41k = 27 + 14k   ………… (1)

We shall prove that P(k + 1) is true whenever P(k) is true

replacing k by k + 1 we have

     41k + 1 – 14k + 1

=  41k . 41 – 14k . 14

=  (27 + 14k) 41 – 14k . 14

=  27 . 41 + 14k .41 – 14k . 14

=  27 . 41 + 14k (41 – 14 )

=  27 . 41 + 14k . 27

=  27 ( 41 + 14k )

 41n – 14n is a multiple of 27
Thus P(k + 1) is true, whenever  P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n N.

Q24. (2n + 7) < (n + 3)2

Solution: Let the statement be p(n) so,

p(n) : (2n + 7) < (n + 3)2

=> p(1) :  (2 × 1 + 7) < (1 + 3)2

=> 9 < 42

=> 9 < 16 

Therefore, p(1) is true so Assume that p(k) is also true for some integer k. 

(2k + 7) < (k + 3)2  ......... (i) 

Now we shall prove for p(k + 1) 

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

2k + 2 + 7 < (k + 4)2  ........ (ii) 

We have from (i) 

(2k + 7) < (k + 3)2 

Adding 2 both sides

=> 2k + 7 + 2 < (k + 3)2 + 2 

=> 2k + 7 + 2 < k2 + 6k + 9 + 2 

=> 2k + 7 + 2 < k2​ + 6k + 9 + 2 

=> 2k + 7 + 2 < k2​ + 6k + 11

Now, k2​ + 6k + 11 < (k + 4)2 from (ii) 

=> 2k + 7 + 2 < k+ 6k + 11 < k2 + 8k + 16 

=> 2k + 2 + 7 < k2 + 8k + 16

=> 2(k + 1) + 7 < (k + 4)2

=> 2(k + 1) + 7 < (k + 1 + 3)2

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

Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n ∈ N.

Other Pages of this Chapter: 4. Principle Of Mathematical Induction

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Study Materials List:

Solutions ⇒ Class 11th ⇒ Mathematics
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle Of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability

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