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

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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 n² + n.
Solution:
Let P(n) : 2 + 4 + 6 +…+2n = n² + n, ∀ n ∈ N

Step 1:
P( 1) : 2 = 1² + 1 = 2 which is true for P( 1)

Step 2:
P(k): 2 + 4 + 6+ …+ 2k = k² + k. Let it be true.

Step 3:
P(k + 1) : 2 + 4 + 6 + … + 2k + (2k + 2)
= k²+ k + (2k + 2) = k² + 3k + 2
= k² + 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, x²ⁿ – y²ⁿ is divisible by x +y.
Solution:
Let P(n) = x²ⁿ – y²ⁿ 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^2ky² + x^2ky² – y^{2k + 2}
= x^2k.x² – x^2ky² + x^2ky² – y^2ky²
= x^2k (x² – y²) + y²λ (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 n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Solution:
Let P(n) : n³ – 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) : k³ – 7k + 3 = 3λ. Let it be true
⇒ k³ = 3λ + 7k – 3

Step 3:
P(k + 1) = (k + 1)³ – 7(k + 1) + 3
= k³ + 1 + 3k² + 3k – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= (3λ + 7k – 3) + 3k² – 4k – 3 (from Step 2)
= 3k² + 3k + 3λ – 6
= 3(k² + 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 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.
Solution:
P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9 = 5² + 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 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 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 + 6²) = 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 10ⁿ + 3 × 4^{n + 2} + 5, is divisible by 9, for all natural numbers n.
Solution:
P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9
P(1) = 10¹ + 3 × 4² + 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 + kx²
⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx² ….. (1)
Now, 1 + (k + 1) x + kx² ≥ 1 + (k + 1)x …… (2)
[∵ kx² > 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.
3²ⁿ – 1 is divisible by 8.
Solution:
P(n) = 3²ⁿ – 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.3² – 1
= 3²(3^2k – 1) + 8
Now, 3^2k – 1 is divisible by 9. [Using (1)]
∴ 3² (3^2k – 1) + 8 is also divisible by 8.
Hence, 3²ⁿ – 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 xⁿ – yⁿ is divisible by x – y. [OR]
xⁿ – yⁿ is divisible by x – y, where x – y ≠ 0.
Solution:
Let the given statement be P(n).
(i.e.) P(n): xⁿ – yⁿ = M(x – y), x – y ≠ 0

Step I.
When n = 1,
xⁿ – yⁿ = 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^ky + 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.) 3⁴ – 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.
n³ – n is divisible by 6, for each natural number n ≥ 2.
Solution:
Let P(n) : n³ – n

Step 1 :
P(2): 2³ – 2 = 6 which is divisible by 6. So it is true for P(2).

Step 2 :
P(A): k³ – k = 6λ. Let it is be true for k ≥ 2
⇒ k³ = 6λ + k …(i)

Step 3 :
P(k + 1) = (k + 1)³ – (k + 1)
= k³ + 1 + 3k² + 3k – k – 1 = k³ – k + 3(k² + k)
= k³ – k + 3(k² + k) = 6λ + k – k + 3(k² + k)
= 6λ + 3(k² + k) [from (i)]
We know that 3(k² + 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, 7ⁿ – 2ⁿ is divisible by 5.
Solution:
Let P(n) : 7ⁿ – 2ⁿ

Step 1 :
P(1) : 7¹ – 2¹ = 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.
n² < 2ⁿ, for all natural numbers n ≥ 5.
Solution:
Let P(n) : n² < 2ⁿ for all natural numbers, n ≥ 5

Step 1 :
P(5) : 1⁵ < 2⁵ ⇒ 1 < 32 which is true for P(5)

Step 2 :
P(k): k² < 2k. Let it be true for k ∈ N

Step 3 :
P(k + 1): (k + 1)² < 2^{k + 1}
From Step 2, we get k² < 2^k
⇒ k² < 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) = 2k² – k + 4k + 1
= 2k² + 3k + 1 = 2k² + 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.