NCERT Solutions for Class 11 Maths Principle of Mathematical Induction [Old Book]

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 4
Chapter Name : Principle of Mathematical Induction

Exercise 4.1

Q1 Prove the following by using the principle of mathematical induction for all n ∈ N:
n
(3 −1)
2 a−1
1 + 3 + 3 + … + 3 =
2

Answer.
Let the given statement be P (n), i.e.,
n
(3 −1)
2 n−1
P (n) : 1 + 3 + 3 + … + 3 =
, which is true
2

′
(3 −1)
3−1 2
P (1) : 1 = = = = 1
2 2 2

Let P (k) be true for some positive integer k, i.e.,

k
(3 −1)
2 k−1
1 + 3 + 3 + … + 3 =
2

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

2 k−1 (k+1)−1
1 + 3 + 3 + … + 3 + 3

2 k−1 k
= (1 + 3 + 3 + … + 3 ) + 3

t
(3 −1)
k
= + 3
[using (i)]
2

t k
(3 −1)+2.3
=
2
k
(1+2)3 −1
=
2
k
3.3 − 1
=
2
k+1
3 − 1
=
2
Thus, P (k + 1) is true whenever P (k) is true.

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

numbers l.e.,

Page : 94 , Block Name : Exercise 4.1

Q2 Prove the following by using the principle of mathematical induction for all n ∈ N:
2
n(n+1)
3 3 3 3
1 + 2 + 3 + … + n = ( )
2

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

2
n(n+1)
3 3 3 3
1 + 2 + 3 + … + n = ( )
2

2 2
3 1.2 2
P (1) : 1 = 1 = ( 1(1 + 1) = ( ) = 1 = 1
2

, which is true.
Let P (k) be true for some positive integer k, l.e.,

2
k(k+1)
3 3 3 3
1 + 2 + 3 + …… + k = ( )
2

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

3 3 3 3 3
1 + 2 + 3 + … + k + (k + 1)
2
k(k+1)
3
= ( ) + (k + 1) [ Using (i)]
2

2 2
k (k + 1)
3
= + (k + 1)
4
2 2 3
k (k + 1) + 4(k + 1)
=
4
2 2
(k + 1) {k + 4(k + 1)}
=
4
2 2
(k+1) {k +4k+4}
=
4

2 2
(k+1) (k+2)
=
4

2 2
(k+1) (k+1+1)
=
4
2
3 3 3 (k+1)(k+1+1)
3 3
= (1 + 2 + 3 + … . +k ) + (k + 1) = ( )
2

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

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

numbers l.e.,

Page : 94 , Block Name : Exercise 4.1

Q3 Prove the following by using the principle of mathematical induction for all n ∈ N:

1 1 1 2n
1 + + + … + =
(1+2) (1+2+3) (1+2+3+…n) (n+1)

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

1 1 1 2n
P(n): 1 + + + … + =
1+2 1+2+3 1+2+3+…n n+1

2.1 2
P (1) : 1 = = = 1
1+1 2

Let P (k) be true for some positive integer k, i.e.,

1 +
1+2
1
+ … +
1

1+2+3
+ … +
1

1+2+3+…+k
=
2k

k+1
............(i)
We shall now prove that P(k + 1) is true.

1 1 1 1
1 + + + … + +
1+2 1+2+3 1+2+3+…+k 1+2+2+k+(k+1)

1 1 1 1
= (1 + + + … + ) +
1+2 1+2+3 1+2+3+.k 1+2+3+…+k+(k+1)

2k 1
= + [ Using (i) ]
k+1 1+2+3+…+k+(k+1)

2k 1 n(n+1)
= + [1 + 2 + 3 + … + n = ]
(k+1)(k+1+1)
k+1 2
( )
2

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

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

Page : 94 , Block Name : Exercise 4.1

Q4 Prove the following by using the principle of mathematical induction for all n ∈ N:
n(n+1)(n+2)(n+3)
1.2.3 + 2.3.4 + … + n(n + 1)(n + 2) =
4

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

n(n+1)(n+2)(n+3)
P (n) : 1.2.3 + 2.3.4 + … + n(n + 1)(n + 2) =
4

1(1+1)(1+2)(1+3) 1.23.4
, which is true
P (1) : 1.2 ⋅ 3 = 6 = = = 6
4 4

k(k+1)(k+2)(k+3)
1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) =
4
……………(i)
We shall now prove that P (k + 1) is true.

1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

= {1.2.3 + 2.3.4 + … + k(k + 1)(k + 2)} + (k + 1)(k + 2)(k + 3)

k(k+1)(k+2)(k+3)
= + (k + 1)(k + 2)(k + 3) [ Using (i)]
4

k
= (k + 1)(k + 2)(k + 3) ( + 1)
4

(k+1)(k+2)(k+3)(k+4)
=
4

(k+1)(k+1+1)(k+1+2)(k+1+3)
=
4

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

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

numbers l.e., n .

Page : 94 , Block Name : Exercise 4.1

Q5 Prove the following by using the principle of mathematical induction for all n ∈ N:
n+1
2 3 n (2n−1)3 +3
1.3 + 2.3 + 3.3 + … + n3 =
4

Answer.
Let the given statement be P (n), i.e.,
n+1
(2n−1)3 +3
2 ∘ n
1.3 + 2.3 + 3.3 + … + n3 =
4

P (n) :

1+1 2
(2.1−1)3 +3 3 +3 12
P (1) : 1.3 = 3 = = = = 3
4 4 4

Let P (k) be true for some positive integer k, i.e.,
k+1

……….(i)
2 3 4 (2k−1)3 +3
1.3 + 2.3 + 3.3 + … + k3 =
4

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

2 3 k k+1
1.3 + 2.3 + 3.3 + … + k3 + (k + 1)3

2 3 k k+1
= (1.3 + 2.3 + 3.3 + … + k.3 ) + (k + 1)3

k+1
(2k−1)3 +3 k+1
= + (k + 1)3 [ Using (i)]
4
3+1 k+1
(2k−1)3 +3+4(k+1)3
=
4

t+1
3 {2k−1+4(k+1)}+3
=
4
k+1
3 {6k + 3} + 3
=
4
k+1
3 ⋅ 3{2k + 1} + 3
=
4
(k+1)+1
3 {2k+1}+3
=
4

(k+1)+1
{2(k+1)−1}3 +3
=
4

Page : 94 , Block Name : Exercise 4.1

Q6 Prove the following by using the principle of mathematical induction for all n ∈ N:
n(n+1)(n+2)
1.2 + 2.3 + 3.4 + … + n. (n + 1) = [ ]
3

n(n+1)(n+2)
P(n) : 1.2 + 2.3 + 3.4 + … + n ⋅ (n + 1) = [ 3
]

1(1+1)(1+2)
P(1): 1.2 = 2 = 3
=
1.2.3

3
= 2 which is true.
k(k+1)(k+2)
1.2 + 2.3 + 3.4 + … . +k. (k + 1) = [
3
] ……….(i)
We shall now prove that P(k + 1) is true.
1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2)
= [1.2 + 2.3 + 3.4 + … + k.(k + 1)] + (k + 1).(k + 2)
k(k+1)(k+2)
= + (k + 1)(k + 2) [ Using (i)]
3

k
= (k + 1)(k + 2) ( + 1)
3

(k+1)(k+2)(k+3)
=
3

(k+1)(k+1+1)(k+1+2)
=
3

Page : 94 , Block Name : Exercise 4.1

Q7 Prove the following by using the principle of mathematical induction for all n ∈ N:
2
n(4n +6n−1)

1.3 + 3.5 + 5.7 + … + (2n − 1)(2n + 1) =
3

2
n(4n +6n−1)
P(n) : 1.3 + 3.5 + 5.7 + … + (2n − 1)(2n + 1) = 3

2
1(4.1 +6.1−1) 4+6−1 9
, which is true
P (1) : 1.3 = 3 = = = = 3
3 3 3

Let P(k) be true for sorne positive integer k, i.e.,
2
k(4k +6k−1)

1.3 + 3.5 + 5.7 + … . +(2k − 1)(2k + 1) =
3
……..(i)
We shall now prove that P (k + 1) is true.

(1.3 + 3.5 + 5.7 + … + (2k − 1)(2k + 1) + {2(k + 1) − 1}{2(k + 1) + 1}

2
k(4k +6k−1)

= + (2k + 2 − 1)(2k + 2 + 1) [ Using (i) ]
3
2
k(4k +6k−1)
= + (2k + 1)(2k + 3)
3

2
k(4k +6k−1)
2
= + (4k + 8k + 3)
3

2 2
k(4k +6k−1)+3(4k +8k+3)
=
3

3 2 2
4k +6k −k+12k +24k+9
=
3

3 2
4k +18k +93k+9
=
3

3 2 2
4k +14k +9k+4k +14k+9
=
3
2 2
k(4k +14k+9)+1(4k +14k+9)
=
3

2
(k+1)(4k +14k+9)
=
3
2
(k+1){4k +8k+4+6k+6−1}
=
3

2
(k+1){4(k +2k+1)+6(k+1)−1}
=
3

2
(k+1){4(k+1) +6(k+1)−1}
=
3

Page : 94 , Block Name : Exercise 4.1

Q8 Prove the following by using the principle of mathematical induction for all n ∈ N:
2 3 ∘ n+1
1.2 + 2.2 + 3.2 + … + n.2 = (n − 1)2 + 2

2 2 n n+1
P (n) : 1.2 + 2.2 + 3.2 + … + n.2 = (n − 1)2 + 2

1+1
P (1) : 1.2 = 2 = (1 − 1)2 + 2 = 0 + 2 = 2,

1.2 + 2.2
2
+ 3.2
2
+ … + k.2
k
= (k − 1)2
k+1
+ 2… ..(i)
We shall now prove that P(k + 1) is true.
2 3 k k+1
{1.2 + 2.2 + 3.2 + … + k.2 } + (k + 1) ⋅ 2

k+1 k+1
= (k − 1)2 + 2 + (k + 1)2

k+1
= 2 {(k − 1) + (k + 1)} + 2

k+1
= 2 ⋅ 2k + 2

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

(k+1)+1
= {(k + 1) − 1}2 + 2

Page : 94 , Block Name : Exercise 4.1

Q9 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 1 1 1 1
+ + + … + n
= 1 − n
2 4 8 2 2

P(n): + + + … +
1

2
1
= 1 −
4
1

8
1

2
n
1

2
n

P(1): = 1 − 1

2
=
1

1
1

2
, which is true
2

1 1 1 1 1
+ + + … + = 1 −
8 k k
2 4 2 2

We shall now prove that P(k + 1) is true.
1 1 1 1 1
( + + + …… + ) +
2 4 8 k k+1

[using(i))]
2 2

1 1
= (1 − ) +
k k+1
2 2

1 1
= 1 − +
k k
2 2⋅2

1 1
= 1 − (1 − )
k 2
2

1 1
= 1 − ( )
k 2
2

1
= 1 −
k+1
2

Page : 94 , Block Name : Exercise 4.1

Q10 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 1 1 1 n
+ + + … + =
2.5 5.8 8.11 (3n−1)(3n+2) (6n+4)

P(n): +
1
+
2.5
+ … +
5.8
1
=
1

8.11 (3n−1)(3n+2)
1 n

(6n+4)

P (1) =
2.5
1
=
1

10
=
6.1+4
1
=
1

10
, which is true.
….(i)
1 1 1 1 k
+ + + … + =
2.5 5.8 8.11 (3k−1)(3k+2) 6k+4

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

1 1 1 1 1
+ + + …… + +
2.5 5.8 8.11 (3k−1)(3k+2) {3(k+1)−1}{3(k+1)+2}

k 1
= + [ Using (i)]
6k+4 (3k+3−1)(3k+3+2)

k 1
= +
6k+4 (3k+2)(3k+5)

k 1
= +
2(3k+2) (3k+2)(3k+5)

1 k 1
= ( + )
(3k+2) 2 3k+5

1 k(3k+5)+2
= ( )
(3k+2) 2(3k+5)

2
1 3k +5k+2
= ( )
(3k+2) 2(3k+5)

1 (3k+2)(k+1)
= ( )
(3k+2) 2(3k+5)

(k+1)
=
6k+10

(k+1)
=
6(k+1)+4

Page : 94 , Block Name : Exercise 4.1

Q11 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 1 1 1 n(n+3)
+ + + … + =
1.2.3 2.3.4 3.4.5 n(n+1)(n+2) 4(n+1)(n+2)

Answer. Let the given statement be P(n),i.e., P(n):
n(n+3)
1 1 1 1
+ + + … + =
1.2.3 2.3.4 3.4.5 n(n+1)(n+2) 4(n+1)(n+2)

1⋅(1+3)
P (1) :
1⋅2⋅3
1
=
4(1+1)(1+2)
=
1⋅4

4⋅2⋅3
=
1

1⋅2⋅3
, which is true
k(k+3)
1

1⋅2⋅3
+
2⋅3⋅4
1
+
1

3⋅4⋅5
+ … +
k(k+1)(k+2)
1
=
4(k+1)(k+2)
…..(i)
We shall now prove that P(k + 1) is true.
1 1 1 1 1
[ + + + … + ] +
1⋅2⋅3 2⋅3⋅4 3⋅4⋅5 k(k+1)(k+2) (k+1)(k+2)(k+3)

k(k+3)
1
= + [ Using (i)]
4(k+1)(k+2) (k+1)(k+2)(k+3)

k(k+3)
1 1
= { + }
(k+1)(k+2) 4 k+3

2
1 k(k+3) +4
= { }
(k+1)(k+2) 4(k+3)

2
k(k +6k+9)+4
1
= { }
(k+1)(k+2) 4(k+3)

3 2
1 k +6k +9k+4
= { }
(k+1)(k+2) 4(k+3)

3 2 2
1 k +2k +k+4k +8k+4
= { }
(k+1)(k+2) 4(k+3)

2 2
k(k +2k+1)+4(k +2k+1)
1
= { }
(k+1)(k+2) 4(k+3)

2 2
1 k(k+1) +4(k+1)
= { }
(k+1)(k+2) 4(k+3)

2
(k+1) (k+4)
=
4(k+1)(k+2)(k+3)

(k+1){(k+1)+3}
=
4{(k+1)+1}{(k+1)+2}

Page : 94 , Block Name : Exercise 4.1

Q12 Prove the following by using the principle of mathematical induction for all n ∈ N:
n
a(r −1)
2 n−1
a + ar + ar + … + ar =
r−1

n
a(r −1)
P(n): a + ar + ar 2
+ … + ar
n−1
=
r−1

1
a(r −1)
P(1) : a =
(r−1)
= a , which is true.
k
a(r −1)
a + ar + ar
2
+ … … . +ar
k−1
=
r−1
…….(i)
We shall now prove that P(k + 1) is true.
2 k−1 (k+1)−1
{a + ar + ar + … … + ar } + ar

k
a(r −1)
[ Using(i) ]
k
= + ar
r−1

k k
a (r − 1) + ar (r − 1)
=
r − 1
k k+1 k
a (r − 1) + ar − ar
=
r − 1
k k+1 k
ar −a+ar −ar
=
r−1

k+1
ar −a
=
r−1

k+1
a(r −1)

=
r−1

Page : 95 , Block Name : Exercise 4.1

Q13 Prove the following by using the principle of mathematical induction for all n ∈ N:
3 5 7 (2n+1)
2
(1 + ) (1 + ) (1 + ) … (1 + ) = (n + 1)
1 4 9 n2

Answer.Let the given statement be P(n), i.e.,
3 5 7 (2n+1)
2
P(n) : (1 + ) (1 + ) (1 + ) … (1 + ) = (n + 1)
9 2
1 4 n

3 2 2
P(1) : (1 + ) = 4 = (1 + 1) = 2 = 4,
1

(2k+1)
(1 +
3

1
) (1 +
5

4
) (1 +
7

9
) … (1 +
2
) = (k + 1)
2
……(1)
k

We shall now prove that P(k + 1) is true.
3 5 7 (2k+1) {2(k+1)+1}
[(1 + ) (1 + ) (1 + ) … (1 + )] {1 + }
9 2 2
1 4 k (k+1)

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

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

2
= {(k + 1) + 1}

Page : 95 , Block Name : Exercise 4.1

Q14 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 1 1 1
(1 + ) (1 + ) (1 + ) … (1 + ) = (n + 1)
1 2 3 n

P(n): (1 + 1

1
) (1 +
1

2
) (1 +
1

3
) … (1 +
1

n
) = (n + 1)

P(1) : (1 +
1

1
) = 2 = (1 + 1) , which is true
P(k) : (1 +
1

1
) (1 +
1

2
) (1 +
1

3
) … (1 +
1

k
) = (k + 1) ……….(1)
We shall now prove that P(k + 1) is true.
1 1 1 1 1
[(1 + ) (1 + ) (1 + ) … (1 + )] (1 + )
1 2 3 k k+1

= (k + 1) (1 +
1

k+1
) [ Using(1) ]
(k+1)+1
= (k + 1) ( )
(k+1)

= (k + 1) + 1

Page : 95 , Block Name : Exercise 4.1

Q15 Prove the following by using the principle of mathematical induction for all n ∈ N:
n(2n−1)(2n+1)
2 2 2 2
1 + 3 + 5 + … + (2n − 1) =
3

2 2 2 n(2n−1)(2n+1)
2
P (n) = 1 + 3 + 5 + … + (2n − 1) =
3

2 1(2.1−1)(2.1+1) 1.1.3
P (1) = 1 = 1 = = = 1, which is true.
3 3

k(2k−1)(2k+1)
P (k) = 1
2
+ 3
2
+ 5
2
+ … + (2k − 1)
2
=
3
…….(1)
We shall now prove that P(k + 1) is true.
2 2 2 2 2
{1 + 3 + 5 + … + (2k − 1) } + {2(k + 1) − 1}

k(2k−1)(2k+1)
=
3
+ (2k + 2 − 1)
2
[ Using (1)]
k(2k−1)(2k+1)
2
= + (2k + 1)
3

2
k(2k−1)(2k+1)+3(2k+1)
=
3

(2k+1){k(2k−1)+3(2k+1)}
=
3

2
(2k+1){2k −k+6k+3}
=
3
2
(2k+1){2k +5k+3}
=
3

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

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

(2k+1)(k+1)(2k+3)
=
3

(k+1){2(k+1)−1}{2(k+1)+1}
=
3

Page : 95 , Block Name : Exercise 4.1

Q16 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 1 1 1 n
+ + + … + =
1.4 4.7 7.10 (3n−2)(3n+1) (3n+1)

1 1 1 1 n
P (n) : + + + … + =
1.4 4.7 7.10 (3n−2)(3n+1) (3n+1)

1 1 1 1
P (1) = = = = , which is true.
1.4 3.1+1 4 1.4

………..(1)
1 1 1 1 k
P (k) = + + + … + =
1.4 4.7 7.10 (3k−2)(3k+1) 3k+1

We shall now prove that P(k + 1) is true.
1 1 1 1 1
{ + + + … + } +
1.4 4.7 7.10 (3k−2)(3k+1) {3(k+1)−2}{3(k+1)+1}

=
k
+
1
[ Using (1)]
3k+1 (3k+1)(3k+4)

1 1
= {k + }
(3k+1) (3k+4)

1 k(3k+4)+1
= { }
(3k+1) (3k+4)

2
1 3k +4k+1
= { }
(3k+1) (3k+4)

2
1 3k +3k+k+1
= { }
(3k+1) (3k+4)

(3k+1)(k+1)
=
(3k+1)(3k+4)

(k+1)
=
3(k+1)+1

Page : 95 , Block Name : Exercise 4.1

Q17 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 1 1 1 n
+ + + … + =
3.5 5.7 7.9 (2n+1)(2n+3) 3(2n+3)

1 1 1 1 n
P (n) : + + + … + =
3.5 5.7 7.9 (2n+1)(2n+3) 3(2n+3)

P (1) :
3.5
1
=
1
=
3.5
1
, which is true.
3(2.1+3)

……(1)
1 1 1 1 k
P (k) : + + + … + =
3.5 5.7 7.9 (2k+1)(2k+3) 3(2k+3)

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

1 1 1 1 1
[ + + + … + ] +
3.5 5.7 7.9 (2k+1)(2k+3) {2(k+1)+1}{2(k+1)+3}

k 1
= + [U sing(1)]
3(2k+3) (2k+3)(2k+5)

1 k 1
= [ + ]
(2k+3) 3 (2k+5)

k(2k+5)+3
1
= [ ]
(2k+3) 3(2k+5)

2
1 2k +5k+3
= [ ]
(2k+3) 3(2k+5)

2
1 2k +2k+3k+3
= [ ]
(2k+3) 3(2k+5)

2k(k+1)+3(k+1)
1
= [ ]
(2k+3) 3(2k+5)

(k+1)(2k+3)
=
3(2k+3)(2k+5)

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

Page : 95 , Block Name : Exercise 4.1

Q18 Prove the following by using the principle of mathematical induction for all n ∈ N:
1 2
1 + 2 + 3 + … + n < (2n + 1)
8

1 2
P(n) : 1 + 2 + 3 + … + n < (2n + 1)
8

It can be noted that P(n) is true for n = 1 since 1 < 1

8
(2.1 + 1)
2
=
9

8

(2k + 1) ……(1)
1 2
1 + 2 + … + k <
8

We shall now prove that P(k + 1) is true whenever P(k) is true.
1 2
(1 + 2 + … + k) + (k + 1) < (2k + 1) + (k + 1) [ Using (1)]
g

1 2
< {(2k + 1) + 8(k + 1)}
8

1 2
< {4k + 4k + 1 + 8k + 8}
8

1 2
< {4k + 12k + 9}
8

1 2
< (2k + 3)
8

1 2
< {2(k + 1) + 1}
8

Hence, (1 + 2 + 3 + … + k) + (k + 1) <
1 2
(2k + 1) + (k + 1)
8

Page : 95 , Block Name : Exercise 4.1

Q19 Prove the following by using the principle of mathematical induction for all n ∈ N:
n(n + 1)(n + 5) is a multiple of 3

P(n): n (n + 1) (n + 5), which is a multiple of 3.
It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.
k (k + 1) (k + 5) is a multiple of 3.
∴k (k + 1) (k + 5) = 3m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
(k + 1){(k + 1) + 1}{(k + 1) + 5}

= (k + 1)(k + 2){(k + 5) + 1}

= (k + 1)(k + 2)(k + 5) + (k + 1)(k + 2)

= {k(k + 1)(k + 5) + 2(k + 1)(k + 5)}

= 3m + (k + 1){2(k + 5) + (k + 2)}

= 3m + (k + 1){2k + 10 + k + 2}

= 3m + (k + 1){2k + 12)

= 3m + 3(k + 1)(k + 4)

= 3{m + (k + 1)(k + 4)} = 3 × q, where q = {m + (k + 1)(k + 4)} is some natural number

Therefore, (k + 1){(k + 1) + 1}{(k + 1) + 5} is a multiple of 3.

Page : 95 , Block Name : Exercise 4.1

Q20 Prove the following by using the principle of mathematical induction for all n ∈ N:
1n−1
10 + 1 is divisible by 11

2n−1
P(n) : 10 + 1 is divisible by 11.

2.1−1
It can be observed that P(n) is true for n = 1 since P(1) = 10 + 1 =

11, which is divisible by 11.

2k−1
10 + 1 is divisible by 11.

2k−1
∴ 10 + 1 = 11m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

2(k+1)−1
10 + 1

2k+2−1
= 10 + 1

2k+1
= 10 + 1

2 2k−1
= 10 (10 + 1 − 1) + 1

2 2k−1 2
= 10 (10 + 1) − 10 + 1

2
= 10 ⋅ 11m − 100 + 1 [ Using (1)]

Page : 95 , Block Name : Exercise 4.1

Q21 Prove the following by using the principle of mathematical induction for all n ∈ N:
2x 3n
x − y is divisible by x + y

2n 2n
P (n) : x − y is divisible by x + y .

It can be observed that P (n) is true for n = 1 .

2 2 2 2
This is so because x × 1 − y × 1 = x − y = (x + y)(x − y) is divisible by

(x + y).

2k 2k
x − y is divisible by x + y.

2k 2k
∴ x − y = m(x + y), where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.
2(k+1) 2(k+1)
x − y

2k 2 2k 2
= x ⋅ x − y ⋅ y

2 2k 2k 2k 2k 2
= x (x − y + y ) − y ⋅ y

2 2k 2k 2
= x {m(x + y) + y } − y ⋅ y [ Using (1)]
2 2k 2 2k 2
= m(x + y)x + y ⋅ x − y ⋅ y

2 2k 2 2
= m(x + y)x + y (x − y )

2 2k
= m(x + y)x + y (x + y)(x − y)

2 2k
= (x + y) {mx + y (x − y)} , which is a factor of (x + y)

Page : 95 , Block Name : Exercise 4.1

Q22 Prove the following by using the principle of mathematical induction for all n ∈ N:
2n+2
3 − 8n − 9 is divisible by 8

2n+2
P(n) : 3 − 8n − 9 is divisible by 8
2×1+2
It can be observed that P (n) is true for n = 1 since 3 − 8 × 1 − 9 =

64, which is divisible by 8 .

2k+2
3 − 8k − 9 is divisible by 8 .

2k+2
∴ 3 − 8k − 9 = 8m; where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.
2(k+1)+2
3 − 8(k + 1) − 9

2k+2 2
= 3 ⋅ 3 − 8k − 8 − 9

2 2k+2
= 3 (3 − 8k − 9 + 8k + 9) − 8k − 17

2 2k+2 2
= 3 (3 − 8k − 9) + 3 (8k + 9) − 8k − 17

= 9.8m + 9(8k + 9) − 8k − 17

= 9.8m + 9(8k + 9) − 8k − 17

= 9.8m + 72k + 81 − 8k − 17

= 9.8m + 64k + 64

= 8(9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

2(k+1)+2
Therefore, 3 − 8(k + 1) − 9 is divisible by 8 .

Page : 95 , Block Name : Exercise 4.1

Q23 Prove the following by using the principle of mathematical induction for all n ∈ N:
n n
41 − 14 is a multiple of 27

n n
P(n) : 41 − 14 is a multiple of 27

It can be observed that P(n) is true for n = 1 since 41 1
− 14
1
= 27 which is a multiple of 27.
k k
41 − 14 is a multiple of 27

k k
∴ 41 − 14 = 27m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

k+1 k+1
41 − 14

k k
= 41 ⋅ 41 − 14 ⋅ 14

k k k k
= 41 (41 − 14 + 14 ) − 14 ⋅ 14

k k k k
= 41 (41 − 14 ) + 41.14 − 14 ⋅ 14

k
= 41.27m + 14 (41 − 14)
k
= 41.27m + 27.14

k
= 27 (41m − 14 )

k
= 27 × r, where r = (41m − 14 ) is a natural number

k+1 k+1
Therefore, 41 − 14 is a multiple of 27

Page : 95 , Block Name : Exercise 4.1

Q24 Prove the following by using the principle of mathematical induction for all n ∈ N:
2
(2n + 7) < (n + 3)

2
P(n) : (2n + 7) < (n + 3)
2
It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)

= 16, which is true.

2
(2k + 7) < (k + 3) … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.
{2(k + 1) + 7} = (2k + 7) + 2
2
∴ {2(k + 1) + 7} = (2k + 7) + 2 < (k + 3) + 2 [ using (1)]
2
2(k + 1) + 7 < k + 6k + 9 + 2

2
2(k + 1) + 7 < k + 6k + 11

2 2
Now, k + 6k + 11 < k + 8k + 16

2
∴ 2(k + 1) + 7 < (k + 4)

2
2(k + 1) + 7 < {(k + 1) + 3}

Page : 95 , Block Name : Exercise 4.1