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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4

**11th Maths Exercise 4.4 Question 1.**

By the principle of mathematical induction, prove that, for n ≥ 1

Solution:

∴ P(k+ 1) is true.

Thus P(K) is true ⇒ (k + 1) is true.

Hence by principle of mathematical induction, P(n) is true for all n ∈ N.

**11th Maths Chapter 4 Exercise 4.4 Question 2.**

By the principle of mathematical induction, prove that, for n > 1

Solution:

∴ P(1) is true

Let P(n) be true for n = k

∴ P(k + 1) is true

Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(k) is true for all n ∈ N.

**Exercise 4.4 Class 11 Question 3.**

Prove that the sum of the first n non-zero even numbers is n^{2} + n.

Solution:

Let P(n) : 2 + 4 + 6 +…+2n = n^{2 }+ n, ∀ n ∈ N

Step 1:

P( 1) : 2 = 1^{2} + 1 = 2 which is true for P( 1)

Step 2:

P(k): 2 + 4 + 6+ …+ 2k = k^{2 }+ k. Let it be true.

Step 3:

P(k + 1) : 2 + 4 + 6 + … + 2k + (2k + 2)

= k^{2}+ k + (2k + 2) = k^{2 }+ 3k + 2

= k^{2 }+ 2k + k + 1 + 1

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

Which is true for P(k + 1)

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

**Ex 4.4 Class 11 Question 4.**

By the principle of Mathematical induction, prove that, for n ≥ 1.

Solution:

∴ P(k + 1) is true

Thus P(k) is true ⇒ P(k + 1) is true

Hence by principle of mathematical induction, P(n) is true for all n ∈ N

**Exercise 4.4 Maths Class 11 Solutions Question 5.**

Using the Mathematical induction, show that for any natural number n ≥ 2,

Solution:

⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true.

**Exercise 4.4 Maths Class 11 Question 6.**

Using the Mathematical induction, show that for any natural number n ≥ 2,

Solution:

⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true for n ≥ 2.

**Ex 4.4 Class 11 Maths Question 7.**

Using the Mathematical induction, show that for any natural number n

Solution:

∴ P(k + 1) is true

Thus p(k) is true ⇒ P(k + 1) is true

Hence by principle of mathematical induction,

p(n) is true for all n ∈ z

**4.4 Maths Class 11 Question 8.**

Using the Mathematical induction, show that for any natural number n,

Solution:

∴ P(k + 1) is true

Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(n) is true for all n ∈ N.

**Exercise 4.4 Class 11 Pdf Question 9.**

Prove by Mathematical Induction that

1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1

Solution:

P(n) is the statement

1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1

To prove for n = 1

LHS = 1! = 1

RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1

LHS = RHS ⇒ P(1) is true

Assume that the given statement is true for n = k

(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true

To prove P(k + 1) is true

p(k + 1) = p(k) + t_{(k + 1)}

P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!

= (k + 1)! + (k + 1) (k + 1)! – 1

= (k + 1)! [1 + k + 1] – 1

= (k + 1)! (k + 2) – 1

= (k + 2)! – 1

= (k + 1 + 1)! – 1

∴ P(k + 1) is true ⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.

**Combinatorics And Mathematical Induction Question 10.**

Using the Mathematical induction, show that for any natural number n, x^{2n} – y^{2n} is divisible by x +y.

Solution:

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

For n = 1

P(1) = x^{2 × 1} – y^{2 × 1} is divisible by (x + y)

⇒ (x + y) (x – y) is divisible by (x + y)

∴ P(1) is true

Let P(n) be true for n = k

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

⇒ x^{2k} – y^{2k} = λ(x + y) …… (i)

For n = k + 1

⇒ P(k + 1) = x^{2(k + 1)} – y^{2(k + 1)} is divisible by (x + y)

Now x^{2(k + 2)} – y^{2(k + 2)}

= x^{2k + 2} – x^{2k}y^{2} + x^{2k}y^{2} – y^{2k + 2}

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

= x^{2k} (x^{2} – y^{2}) + y^{2}λ (x + y) [Using (i)]

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

∴ P(k + 1) is true.

Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(n) is true for all n ∈ N

**Chapter 4 Class 11 Maths Question 11.**

By the principle of mathematical induction, prove that, for n ≥ 1,

Solution:

**Chapter 4 Maths Class 11 Question 12.**

Use induction to prove that n^{3} – 7n + 3, is divisible by 3, for all natural numbers n.

Solution:

Let P(n) : n^{3} – 7n + 3

Step 1:

P(1) = (1)^{3} – 7(1) + 3

= 1 – 7 + 3 = -3 which is divisible by 3

So, it is true for P(1).

Step 2:

P(k) : k^{3} – 7k + 3 = 3λ. Let it be true

⇒ k^{3} = 3λ + 7k – 3

Step 3:

P(k + 1) = (k + 1)^{3} – 7(k + 1) + 3

= k^{3} + 1 + 3k^{2} + 3k – 7k – 7 + 3

= k^{3} + 3k^{2} – 4k – 3

= (3λ + 7k – 3) + 3k^{2} – 4k – 3 (from Step 2)

= 3k^{2} + 3k + 3λ – 6

= 3(k^{2} + k + λ – 2) which is divisible by 3.

So it is true for P(k + 1).

Hence, P(k + 1) is true whenever it is true for P(k).

**10th Maths Exercise 4.4 11th Sum Question 13.**

Use induction to prove that 5^{n + 1} + 4 × 6^{n} when divided by 20 leaves a remainder 9, for all natural numbers n.

Solution:

P(n) is the statement 5^{n + 1} + 4 × 6^{n} – 9 is ÷ by 20

P(1) = 5^{1 + 1} + 4 × 6^{1} – 9 = 5^{2} + 24 – 9

= 25 + 24 – 9 = 40 ÷ by 20

So P(1) is true

Assume that the given statement is true for n = k

(i.e) 5^{k + 1} + 4 × 6^{n} – 9 is ÷ by 20

P(1) = 5^{1 + 1} + 4 × 6^{1} – 9

= 25 + 24 – 9

So P(1) is true

To prove P(k + 1) is true

P(k + 1) = 5^{k + 1 + 1} + 4 × 6^{k + 1 + 1} – 9

= 5 × 5 ^{k + 1} + 4 × 6 × 6^{k} – 9

= 5[20C + 9 – 4 × 6^{k}] + 24 × 6^{k} – 9 [from(1)]

= 100C + 45 – 206^{k} + 246^{k} – 9

= 100C + 46^{k} + 36

= 100C + 4(9 + 6^{k})

Now for k = 1 ⇒ 4(9 + 6^{k}) = 4(9 + 6)

= 4 × 15 = 60 ÷ by 20 .

for k = 2 = 4(9 + 6^{2}) = 4 × 45 = 180 ÷ 20

So by the principle of mathematical induction 4(9 + 6^{k}) is ÷ by 20

Now 100C is ÷ by 20.

So 100C + 4(9 + 6^{k}) is ÷ by 20

⇒ P(k + 1) is true whenever P(k) is true. So by the principle of mathematical induction P(n) is true.

**Samacheer Kalvi 11th Maths Question 14.**

Use induction to prove that 10^{n} + 3 × 4^{n + 2} + 5, is divisible by 9, for all natural numbers n.

Solution:

P(n) is the statement 10^{n} + 3 × 4^{n + 2} + 5 is ÷ by 9

P(1) = 10^{1} + 3 × 4^{2} + 5 = 10 + 3 × 16 + 5

= 10 + 48 + 5 = 63 ÷ by 9

So P(1) is true. Assume that P(k) is true

(i.e.) 10^{k} + 3 × 4^{k + 2} + 5 is ÷ by 9

(i.e.) 10^{k} + 3 × 4^{k + 2} + 5 = 9C (where C is an integer)

⇒ 10^{k} = 9C – 5 – 3 × 4^{k + 2} ……(1)

To prove P(k + 1) is true.

Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5

= 10 × 10^{k} + 3 × 4^{k + 2} × 4 + 5

= 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5

= 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5

= 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5

= 90C – 45 – 18 × 4^{k + 2}

= 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9

So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.

Question 15.

Prove that using the Mathematical induction

Solution:

### Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Additional Questions

Question 1.

Prove by induction the inequality (1 + x)^{n} ≥ 1 + nx, whenever x is positive and n is a positive integer.

Solution:

P(n) : (1 +x)^{n} ≥ 1 +nx

P(1): (1 + x)^{1} ≥ 1 + x

⇒ 1 + x ≥ 1 + x, which is true.

Hence, P(1) is true.

Let P(k) be true

(i.e.) (1 + x)^{k} ≥ 1 + kx

We have to prove that P(k + 1) is true.

(i.e.) (1 + x)^{k + 1} ≥ 1 + (k + 1)x

Now, (1 + x)^{k + 1} ≥ 1 + kx [∵ p(k) is true]

Multiplying both sides by (1 + x), we get

(1 + x)^{k}(1 + x) ≥ (1 + kx)(1 + x)

⇒ (1 + x)^{k + 1} ≥ 1 + kx + x + kx^{2}

⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx^{2} ….. (1)

Now, 1 + (k + 1) x + kx^{2} ≥ 1 + (k + 1)x …… (2)

[∵ kx^{2} > 0]

From (1) and (2), we get

(1 + x)^{k + 1} ≥ 1 + (k + 1)x

∴ P(k + 1) is true if P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all values, of n.

Question 2.

3^{2n} – 1 is divisible by 8.

Solution:

P(n) = 3^{2n} – 1 is divisible by 8

For n = 1, we get

P(1) = 3^{2.1} – 1 = 9 – 1 = 8

P(1) = 8, which is divisible by 8.

Let P(n) be true for n = k

P(k) = 3^{2k} – 1 is divisible by 8 ….. (1)

Now, P(k + 1) = 3^{(2k + 2)} – 1 = 3^{2k}.3^{2} – 1

= 3^{2}(3^{2k} – 1) + 8

Now, 3^{2k} – 1 is divisible by 9. [Using (1)]

∴ 3^{2} (3^{2k} – 1) + 8 is also divisible by 8.

Hence, 3^{2n} – 1 is divisible by 8 ∀ n E N

Question 3.

Prove by the principle of mathematical induction if x and y are any two distinct integers, then x^{n} – y^{n} is divisible by x – y. [OR]

x^{n} – y^{n} is divisible by x – y, where x – y ≠ 0.

Solution:

Let the given statement be P(n).

(i.e.) P(n): x^{n} – y^{n} = M(x – y), x – y ≠ 0

Step I.

When n = 1,

x^{n} – y^{n} = x – y = M(x – y) ….(1)

⇒ P(1) is true.

Step II.

Assume that P(k) is true.

(i.e.) x^{k} – y^{k} = M(x – y), x – y ≠ 0

We shall now show that P(k + 1) is true

Now, x^{k + 1} – y^{k + 1} = x^{k + 1} – x^{k}y + x^{k + 1}y – y^{k + 1}

= x^{k}(x – y) + y(x^{k} – y^{k})

= x^{k}(x – y) + yM(x – y) [Usng ….. (1)]

= (x – y)(x^{k} – yM)

∴ By the principle of mathematical induction, P(n) is true for all n ∈ N

Question 4.

Prove by the principle of mathematical induction that for every natural number n, 3^{2n + 2} – 8n – 9 is divisible by 8.

Solution:

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

Then, P(1): 3^{2.1 + 2} – 8.1 – 9 is divisible by 8.

(i.e.) 3^{4} – 8 – 9 is divisible by 8 or 81 – 8 – 9 is divisible by 8

(or) 64 is divisible by 8, which is true.

Suppose P(k) is true, then

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

(i.e.) 3^{2k + 2} – 8k – 9 = 8m, where m ∈ N (or)

3^{2k + 2} = 8m + 8k + 9

P(k + 1) is the statement given by, …(1)

P(k + 1) : 3^{2(k + 1) + 2} – 8(k + 1) – 9

∴ P(k + 1) is true

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 5.

Use the principle of mathematical induction to prove that for every natural number n.

Solution:

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

⇒ P(1) is true.

We note than P(n) is true for n = 1.

Assume that P(k) is true

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

∴ P(k + 1) is also true whenever P(k) is true

Hence, by the principle of mathematical induction, P(n) is also true for all n ∈ N.

Question 6.

n^{3} – n is divisible by 6, for each natural number n ≥ 2.

Solution:

Let P(n) : n^{3} – n

Step 1 :

P(2): 2^{3} – 2 = 6 which is divisible by 6. So it is true for P(2).

Step 2 :

P(A): k^{3} – k = 6λ. Let it is be true for k ≥ 2

⇒ k^{3} = 6λ + k …(i)

Step 3 :

P(k + 1) = (k + 1)^{3} – (k + 1)

= k^{3} + 1 + 3k^{2} + 3k – k – 1 = k^{3} – k + 3(k^{2} + k)

= k^{3} – k + 3(k^{2} + k) = 6λ + k – k + 3(k^{2} + k)

= 6λ + 3(k^{2} + k) [from (i)]

We know that 3(k^{2} + k) is divisible by 6 for every value of k ∈ N.

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

Question 7.

For any natural number n, 7^{n} – 2^{n} is divisible by 5.

Solution:

Let P(n) : 7^{n} – 2^{n}

Step 1 :

P(1) : 7^{1} – 2^{1} = 5λ which is divisible by 5. So it is true for P(1).

Step 2 :

P(k): 7^{k} – 2^{k} = 5λ. Let it be true for P(k)

Step 3 :

P(k + 1) = 7^{k + 1} – 2^{k + 1}

So, it is true for P(k + 1)

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

Question 8.

n^{2} < 2^{n}, for all natural numbers n ≥ 5.

Solution:

Let P(n) : n^{2} < 2^{n} for all natural numbers, n ≥ 5

Step 1 :

P(5) : 1^{5} < 2^{5} ⇒ 1 < 32 which is true for P(5)

Step 2 :

P(k): k^{2} < 2k. Let it be true for k ∈ N

Step 3 :

P(k + 1): (k + 1)^{2} < 2^{k + 1}

From Step 2, we get k^{2} < 2^{k}

⇒ k^{2} < 2^{k + 1} < 2k + 2k + 1

From eqn. (i) and (ii), we get (k + 1)^{2} < 2^{k + 1}

Hence, P(k + 1) is true whenever P(k) is true for k ∈ N, n ≥ 5.

Question 9.

In 2n < (n + 2)! for all natural number n.

Solution:

Let P(n) : 2n < (n + 2)! for all k ∈ N.

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

Question 10.

1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N.

Solution:

Let P(n) : 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N

Step 1:

P(1) : 1 = 1(2.1 – 1) = 1 which is true for P(1)

Step 2:

P(k) : 1 + 5 + 9 + … + (4k – 3) = k(2k – 1). Let it be true.

Step 3:

P(k + 1) : 1 + 5 + 9 + … + (4k – 3) = k(4k + 1)

= k(2k – 1) + (4k + 1) = 2k^{2} – k + 4k + 1

= 2k^{2} + 3k + 1 = 2k^{2} + 2k + k + 1

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

= (k+ 1) (2k + 2 – 1) = (k + 1) [2(k + 1) – 1]

Which is true for P(k + 1).

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