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

Step 1:
Let us verify the statement for n = 1
p(1) = 13 = $$\left(\frac{1(1+1)}{2}\right)^{2}$$

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

Step 3 :
Let us prove the statement is true for n = k + 1
P(k + 1 ) = 13 + 23 + 33 + …………… + k3 + (k + 1)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.

Question 2.
By the principle of mathematical induction, prove that, for n ≥ 1

Step 1 :
Let us verify the statement for n = 1

∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k

Step 3:
Let us prove the statement is true for n = k + 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.

is true for all natural numbers n.

Question 3.
Prove that the sum of the first ’n’ non-zero even numbers is n2 + n.
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, 1 . 2 + 2 . 3 + 3 . 4 + …………. + n . (n + 1) = $$\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}$$
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 ) =

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)

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

Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Let

The first stage is n = 2

Step 1:
Let us verify the result for n = 2

∴ The given result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

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

∴ 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.

is true for all natural numbers n ≥ 2.

Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Step 1:
Since n ≥ 2 the first stage is n = 2
Let us verify the result for n = 2

∴ P (2 ) is true.
The result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

Step 3:
Let us prove the result for n = k + 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 ≥ 2.

is true for all natural numbers n ≥ 2

Question 7.
Using the mathematical induction, show that for any natural number n,

Step 1:
First let us verify the result for n = 1

∴ The given result is true for n = 1

Step 2:
Let us assume the result for n = k

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

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)

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)

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.

is true for all natural numbers n

Question 8.
Using the Mathematical induction, show that for any natural number n,

Step 1:
First let us verify the result for n = 1

∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k

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

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

∴ 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, 12 + 22 + 32 + ………….. + n2 > $$\frac{n^{3}}{3}$$
Let P (n) = 12 + 22 + 32 + ………….. + n2 > $$\frac{n^{3}}{3}$$

Step 1:
Let us first verify the result for n = 1

∴ 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

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 > $$\frac{n^{3}}{3}$$

Question 12.
Use induction to prove that n3 – 7n + 3, is divisible by 3, for all natural numbers n.
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 5n + 1 + 4 × 6n when divided by 20 leaves a remainder 9 , for all natural numbers n.
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

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.

Question 14.
Use induction to prove that 10n + 3 × 4n + 2 + 5 is divisible by 9, for all natural numbers n.
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

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.

Question 15.
Prove that using mathematical induction