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# Exercise 4.1 Class 11 maths 4. Principle Of Mathematical Induction - ncert solutions - Toppers Study

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## Exercise 4.1 Class 11 maths 4. Principle Of Mathematical Induction - ncert solutions - Toppers Study

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### Exercise 4.1 class 11 Mathematics Chapter 4. Principle Of Mathematical Induction

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- Exercise 4.1 Class 11 Maths 4. Principle Of Mathematical Induction - Ncert Solutions - Toppers Study
- Class 11 Ncert Solutions
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- Chapter 4. Principle Of Mathematical Induction Exercise 4.1 Class 11

## 4. Principle Of Mathematical Induction

### | Exercise 4.1 |

## Exercise 4.1 Class 11 maths 4. Principle Of Mathematical Induction - ncert solutions - Toppers Study

**Chapter 4. Principle of Mathematical induction **

**Exercise 4.1 **

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

= k^{3} + 6k^{2} + 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) (k^{2} + 8k + 12)

= k (k^{2} + 8k + 12) + 1(k^{2} + 8k + 12)

= k^{3} + 8k^{2} + 12k + k^{2} + 8k + 12

= k^{3} + 9k^{2} + 20k + 12

=( k^{3} + 6k^{2} + 5 k) + 3k^{2} + 15k + 12

= 3m + 3k^{2} + 15k + 12 from (1)

= 3(m + k^{2} + 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. 10 ^{2n - 1 } + 1 is divisible by 11. **

**Solution: **

Let the given statement be P(n), so

P(n) : 10^{2n - 1 } + 1 is divisible by 11.

For n = 1, so we have;

10^{2n - 1 } + 1 = 10^{2}^{×}^{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.

10^{2k- 1 } + 1 = 11m say

10^{2k- 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

10^{2k - 1 } + 1

= 10^{2k + 1 } + 1

= 10^{2k} × 10^{1} + 1

= {10^{2k - 1} × 100 + 1}

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

= 1100m - 100+ 1

= 1100m - 99

= 11(100m - 9)

∴ 10^{2n - 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. ***x*^{2n}**– y^{2n} is divisible by x + y**

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

P(n) : x^{2n} – y^{2n} is divisible by x + y

Putting n = 1 we have,

x^{2n} – y^{2n} = x^{2} - y^{2} = (x + y) (x - y)

P(n) is true for n = 1

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

x^{2k} – y^{2k} is divisible by (x + y)

So, x^{2k} – y^{2k} = m( x + y)

Or x^{2k} = m( x + y) + y^{2k} …………. (1)

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

∴ replacing k by k + 1 we have

x^{2k + 2} – y^{2k + 2}

= x^{2k }. x^{2} – y^{2k }.y^{2}

Putting the value of x^{2k} from (1)

= {m( x + y) + y^{2k}} x^{2} – y^{2k }.y^{2}

= m( x + y) x^{2} + y^{2k}. x^{2} – y^{2k }.y^{2}

= m( x + y) x^{2} + y^{2k} (x^{2} – y^{2})

= m( x + y) x^{2} + y^{2k} (x + y) ( x - y)

= ( x + y) [mx^{2} + y^{2k} ( x - y)]

∴ x^{2n} – y^{2n} 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. ****3 ^{2n+2} – 8n – 9 is divisible by 8**

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

P(n) : 3^{2n+2} – 8*n *– 9 is divisible by 8

Putting n =1

P(1) : 3^{2×1+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

3^{2k + 2} – 8*k *– 9

3^{2k + 2} – 8*k *– 9 is divisible by 8

3^{2k + 2} – 8*k *– 9 = 8m

Or 3^{2k + 2} = 8m + 8*k *+ 9 ……………. (1)

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

∴ replacing k by k + 1 we have

3^{2k + 4} – 8*k *–* 8 *– 9

= 3^{2k + 4} – 8*k *–* 17*

= 3^{2k + 2} × 3^{2} – 8*k *–* 17*

= (8m + 8*k *+ 9)× 9 – 8*k *–* 17*

= 72m + 72*k *+ 81 – 8*k *–* 17*

= 72m + 64*k *+ 64

= 8(9m + 8*k *+ 8)

∴ 3^{2n+2} – 8*n *– 9 is divisible by 8

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

*n*) is true for all natural numbers n ∈ N.

**Q23. ****41^{n} – 14^{n} is a multiple of 27.**

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

P(n) : 41* ^{n} *– 14

*is a multiple of 27*

^{n}Putting n = 1

P(1): 41* ^{n} *– 14

*41 – 14 = 27*

^{n}=∴ P(1) is true

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

41* ^{k} *– 14

*= 27*

^{k}41* ^{k} *= 27 + 14

*(1)*

^{k}…………We shall prove that P(k + 1) is true whenever P(k) is true

∴ replacing k by k + 1 we have

41* ^{k + 1} *– 14

^{k + 1 }= 41* ^{k }. *41 – 14

*. 14*

^{k }= (27 + 14* ^{k}*) 41 – 14

*. 14*

^{k }= 27 . 41 + 14* ^{k} .*41 – 14

*. 14*

^{k }= 27 . 41 + 14* ^{k} *(41 – 14 )

= 27 . 41 + 14* ^{k} *. 27

= 27 ( 41 + 14* ^{k} *)

∴ 41* ^{n} *– 14

*is a multiple of 27*

^{n}Thus P(

*k*+ 1) is true, whenever P(

*k*) is true.

*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 < 4^{2}

=> 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 < k^{2} + 6k + 9 + 2

=> 2k + 7 + 2 < k^{2} + 6k + 9 + 2

=> 2k + 7 + 2 < k^{2} + 6k + 11

Now, k^{2} + 6k + 11 < (k + 4)^{2} from (ii)

=> 2k + 7 + 2 < k^{2 }+ 6k + 11 < k^{2} + 8k + 16

=> 2k + 2 + 7 < k^{2} + 8k + 16

=> 2(k + 1) + 7 < (k + 4)^{2}

=> 2(k + 1) + 7 < (k + 1 + 3)^{2}

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

*n*) is true for all natural numbers n ∈ N.

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

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