**NCERT Solutions For Class 11 Math Chapter -4 Exercise – 4.1 **

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

**1.**

**Solution:**

P

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 + 2.2 ^{2} + 3.2^{2} + … + n.2^{n} = (n – 1) 2^{n}^{+1} + 2**

**Solution:**

We can write the given statement as

P (*n*): 1.2 + 2.2^{2} + 3.2^{2} + … + *n*.2* ^{n}* = (

*n*– 1) 2

^{n}^{+1}+ 2

If we get

P (1): 1.2 = 2 = (1 – 1) 2^{1+1} + 2 = 0 + 2 = 2

Which is true.

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

1.2 + 2.2^{2} + 3.2^{2} + … + *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. 10 ^{2}^{n}^{ – 1 }+ 1 is divisible by 11**

**Solution:**

We can write the given statement as

P (*n*): 10^{2}^{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^{2}^{k}^{ – 1 }+ 1 is divisible by 11

10^{2}^{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} (10^{2k-1} + 1 – 1) + 1

We get

= 10 ^{2} (10^{2k-1} + 1) – 10^{2} + 1

Using equation 1 we get

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

**Solution:**

We can write the given statement as

P (*n*): *x*^{2}* ^{n}* –

*y*

^{2}

*is divisible by*

^{n}*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}

*is divisible by*

^{k}*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} . x^{2} – y^{2k} . y^{2}

By adding and subtracting y^{2k} we get

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

From equation (1) we get

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

By multiplying the terms

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

Taking out the common terms

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

Expanding using formula

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

So we get

= (x + y) {mx^{2} + y^{2k} (x – y)}, which is a factor of (x + y)

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

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

**Solution:**

We can write the given statement as

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

If n = 1 we get

P (1) = 3^{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^{2}^{k}^{ + 2} – 8*k* – 9 is divisible by 8

3^{2}^{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} . 3^{2} – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 3^{2} (3^{2k + 2} – 8k – 9 + 8k + 9) – 8k – 17

On further simplification

= 3^{2} (3^{2k + 2} – 8k – 9) + 3^{2} (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)

= 8 where = (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} – 14^{n} is a multiple of 27**

**Solution:**

We can write the given statement as

P (*n*):41* ^{n}* – 14

*is a multiple of 27*

^{n}If n = 1 we get

P (1) = 41^{1} – 14^{1} = 27, which is a multiple by 27

Which is true.

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

41* ^{k}* – 14

*is a multiple of 27*

^{k}41* ^{k}* – 14

*= 27*

^{k}*m*, 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.

**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 < k^{2} + 6k + 9 + 2

On further calculation

2 (k + 1) + 7 < k^{2} + 6k + 11

Here k^{2} + 6k + 11 < k^{2} + 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.