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

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.

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

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

Step 1:
Let us verify the statement for n = 1
P (1 ) = 2 = 1² + 1 = 1 + 1 = 2.
∴ The given result is true for n = 1.

Step 2:
Let us assume that the given result is true for n = k
P ( k) = 2 + 4 + 6 + ………… + 2k = k² + k

Step 3:
Let us prove the result for n = k + 1
P (k+ 1 ) = 2 + 4 + 6+ + 2k + (2k + 2 )
P(k+ 1 ) = P(k) + (2k + 2)
= k² + k + 2k + 2
= k² + 3k + 2
= k² + 2k + k + 2
= k(k +2) + 1(k + 2)
P(k+ 1 ) = (k+ 1) (k + 2) ……….. (1)
P (k) = k² + k
= k (k + 1 )
P(k+ 1) = (k + 1) (k + 1 + 1)
= (k + 1) (k + 2)
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
2 + 4 + 6 + ………….. + 2n = n² + n
is true for all natural numbers n.

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

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.

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.

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 the principle of mathematical induction,
p(n) is true for all n ∈ z

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

Question 9.
Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1
Solution:
Let p(n) = 1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1

Step 1:
First let us verify the result for n = 1
P(1) = 1! = (1 + 1)! – 1
P(1) = 1! = 2! – 1
P(1) = 1 = 2 – 1 = 1
∴ We have verified the result for n = 1.

Step 2:
Let us assume that the result is true for n = k
P(k) = (1 × 1 !) + (2 × 2!) + (3 × 3!) + …………. + (k × k!) = (k + 1)! – 1

Step 3:
Let us prove the result for n = k + 1
P(k + 1)=(1 × 1!) + (2 × 2!) + (3 × 3!) + ………….. + (k × k!) + ((k + 1) × (k + 1)!)
P(k + 1) = P(k) + ((k + 1) × (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
P(k + 1) = ((k + 1) + 1)! – 1
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
(1 × 1!) + (2 × 2!) + (3 × 3!) + …………… + (n × n!) = (n + 1)! – 1
is true for all natural numbers n.

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
Step 1:
First, let us verify the result for n = 1.
P ( 1 ) = x²⁽¹⁾ – y²⁽¹⁾ = x² – y²
P(1) = (x + y) (x – y) which is divisible by x + y
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P(k) = x^2k – y^2k which is divisible by x + y
∴ P (k) = x^2k – y^2k = λ (x + y) where λ ∈ N ——— (1)

Step 3:
Let us prove the result for n = k + 1

∴ P ( k + 1) is divisible by x + y
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
x²ⁿ – y²ⁿ is divisible by x + y
for all natural numbers n.

Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1,

Solution:

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 is divisible by 3

Step 1:
First let us verify the results for n = 1
P(I) = 1³ – 7 × 1 + 3
= 1 – 7 + 3
P (1) = – 3
which is divisible by 3
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P(k) = k³ – 7k + 3 is divisible by 3
P(k) = k³ – 7k + 3 = 3λ where λ ∈ N

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = (k + 1)³ – 7(k + 1 ) + 3
= k³ + 3k² + 3k + 1 – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= k³ – 4k – 3k + 3k – 3 + 6 – 6 + 3k²
= k³ – 7k + 3 + 3k – 6 + 3k²
= (k³ – 7k + 3) + 3(k² + k – 2)
= 3λ + 3 (k² + k – 2)
P(k + 1) = 3 (λ + k² + k – 2 )
which is a multiple of 3, hence divisible by 3
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
n³ – 7n + 3 is divisible by 3 for all natural numbers n.

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.

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 that 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.