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

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Text Book Back Questions and Answers, Notes.

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
12 + 22 + 32 + ……….. + n3 = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
Answer:
Let P(n) = 12 + 22 + 32 + ……….. + n3
p(n) = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)

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

Step 1:
Let us verify the statement for n = 1
p(1) = 13 = \(\left(\frac{1(1+1)}{2}\right)^{2}\)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 1
P(1) = 1
∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k
P(k) = 13 + 23 + 33 + …………… + k3
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 2

Step 3 :
Let us prove the statement is true for n = k + 1
P(k + 1 ) = 13 + 23 + 33 + …………… + k3 + (k + 1)3
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 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.
12 + 22 + 32 + ……….. + n3 = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
is true for all natural numbers n.

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

Question 2.
By the principle of mathematical induction, prove that, for n ≥ 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 4
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 5

Step 1 :
Let us verify the statement for n = 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 6
∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 7

Step 3:
Let us prove the statement is true for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 8
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 9
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.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 10
is true for all natural numbers n.

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

Question 3.
Prove that the sum of the first ’n’ non-zero even numbers is n2 + n.
Answer:
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.

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

Question 4.
By the principle of mathematical induction, prove that, for n ≥ 1, 1 . 2 + 2 . 3 + 3 . 4 + …………. + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)
Answer:
Let P(n) = 1 . 2 + 2 . 3 + 3 . 4 + ………… + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)

Step 1:
Let us verify the statement for n = 1
P(1) = 1 . 2 = \(\frac{1(1+1)(1+2)}{3}\)
= 1 . 2 = \(\frac{1 \cdot 2 \cdot 3}{3}\)
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P (k) = 1-2 + 2-3 + 3- 4 + + k (k + 1 ) =
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 11

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + …………… + k (k + 1) (k + 1) (k + 1 + 1)
P (k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + ………….. + k (k + 1) + (k + 1 ) (k + 2)
P (k + 1 ) = P(k) + (k + 1) (k + 2)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 12
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 . 2 + 2 . 3 + 3 . 4 + ………. + n (n + 1) = \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}\)
is true for all natural numbers n ≥ 1

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

Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 13
Answer:
Let
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 14
The first stage is n = 2

Step 1:
Let us verify the result for n = 2
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 15
∴ The given result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 16

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 17
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 18
∴ 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.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 19
is true for all natural numbers n ≥ 2.

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

Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 20
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 21

Step 1:
Since n ≥ 2 the first stage is n = 2
Let us verify the result for n = 2
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 22
∴ P (2 ) is true.
The result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 23

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 24
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 25
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.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 26
is true for all natural numbers n ≥ 2

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

Question 7.
Using the mathematical induction, show that for any natural number n,
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 27
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 28

Step 1:
First let us verify the result for n = 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 29
∴ The given result is true for n = 1

Step 2:
Let us assume the result for n = k
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 30

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 31
Factorizing k3 + 6k2 + 9k + 4
f (k) = k3 + 6k2 + 9k + 4
f(- 1) = (- 1)3 + 6(- 1)2 + 9(- 1) + 4
f(- 1) = – 1 + 6 – 9 + 4 = 0
∴ (k + 1) is a factor of f(k)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 32
k3 + 6k2 + 9k + 4 = (k + 1) (k2 + 5k + 4).
= (k + 1) (k2 + 4k + k + 4)
= (k + 1) [k(k + 4) + 1(k + 4)]
= (k + 1) (k + 4) (k + 1)
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 33
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.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 34
is true for all natural numbers n

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

Question 8.
Using the Mathematical induction, show that for any natural number n,
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 35
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 36

Step 1:
First let us verify the result for n = 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 37
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 38
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 39

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 40
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 41
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.
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 42

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

Question 9.
Prove by mathematical induction that
1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1
Answer:
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.

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

Question 10.
Using the Mathematical induction, show that for any natural number n, x2n – y2n is divisible by x + y.
Answer:
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) = x2k – y2k which is divisible by x + y
∴ P (k) = x2k – y2k = λ (x + y) where λ ∈ N ——— (1)

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 43
∴ 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.

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

Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1, 12 + 22 + 32 + ………….. + n2 > \(\frac{n^{3}}{3}\)
Answer:
Let P (n) = 12 + 22 + 32 + ………….. + n2 > \(\frac{n^{3}}{3}\)

Step 1:
Let us first verify the result for n = 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 44
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k.
P(k) = 12 + 22 + 32 + …………… + k2 > \(\frac{\mathrm{k}^{3}}{3}\)

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 12 + 22 + 32 + …………… + k2 + (k + 1)2
P(k + 1) = P(k) + ( k + 1)2
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 45
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.
12 + 22 + 32 + ………….. + n2 > \(\frac{n^{3}}{3}\)

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

Question 12.
Use induction to prove that n3 – 7n + 3, is divisible by 3, for all natural numbers n.
Answer:
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.

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

Question 13.
Use induction to prove that 5n + 1 + 4 × 6n when divided by 20 leaves a remainder 9 , for all natural numbers n.
Answer:
To prove 5n + 1 + 4 × 6n when divided by 20 leaves a remainder 9
That is to prove 5n + 1 + 4 × 6n – 9 is divisible by 20
Let P (n) be the statement 5n + 1 + 4 × 6n
which is divisible by 20
P (n) = 5n + 1 + 4 × 6n – 9 = 20λ

Step 1 :
Let us verify the statement for n = 1
P (1) = 5i + 1 + 4 × 61 – 9
= 52 + 4 × 6 – 9
= 25 + 24 – 9
= 49 – 9 = 40
which is divisible by 20
∴ The statement is true for n = 1

Step 2:
Let us assume that the statement is true for n = k
P(k) = 5k + 1 + 4 × 6k – 9 is divisible by 20
P(k) = 5k + 1 + 4 × 6k – 9 = 20λ …………… (1)

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 46
Hence, P(k + 1) is divisible by 20.
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.
Thus, 5n + 1 + 4 × 6n when divided by 20 leaves a remainder 9, for all natural numbers n.

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

Question 14.
Use induction to prove that 10n + 3 × 4n + 2 + 5 is divisible by 9, for all natural numbers n.
Answer:
Let P( n) = 10n + 3 × 4n + 2 + 5 is divisible by 9

Step 1:
Let us verify the result for n = 1
P ( 1 ) = 101 + 3 × 41 + 2 + 5
= 15 + 3 × 43
= 15 + 3 × 64
= 15 + 192 = 207
which is divisible by 9
Thus we have verified the result for n = 1

Step 2:
Let us assume the result is true for n = k
P(k) = 10k + 3 × 4k + 2 + 5 is divisible by 9
∴ 10k + 3 × 4k + 2 + 5 = 9m for some m
10k = 9m – 5 – 3 × 4k + 2 for some m

Step 3:
Let us prove the result for n = k + 1
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 47
which is divisible by 9
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.
10n + 3 × 4n + 2 + 5 is divisible by 9,
for all natural numbers n.

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

Question 15.
Prove that using mathematical induction
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 48
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 49
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 50

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

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 51
Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 52

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

Samacheer Kalvi 11th Maths Guide Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 53
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.

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