# Question-1

**1+3+3**

^{2}+â€¦ +3^{n-1}= .**Solution:**

Let P(n) = 1+3+3

^{2}+â€¦ +3

^{n-1}

P(1) is true

Assuming P(k) is true

1+3+3

^{2}+â€¦ +3

^{k-1}=

To prove P(k +1) is true using the result of P(k)

1+3+3

^{2}+â€¦ +3

^{k-1}+3k = â€¦â€¦â€¦â€¦â€¦(1)

proof: Taking L.H.S of (1)

1+3+3

^{2}+â€¦ +3

^{k-1}+3k

= + 3k

= =

=

Which is the R.H.S of (1)

P(K + 1 ) is true

By the Principle of Mathematical induction, given expression is true for all values of n .

When n N

Hence proved.

# Question-2

**1**

^{3}+ 2^{3}+ 3^{3}+ â€¦. +n^{3}= .**Solution:**

Let P(n) = 1

^{3}+ 2

^{3}+ 3

^{3}+ â€¦. +n

^{3}=

P(1) is true

Assume P(k) is true.

1

^{3}+ 2

^{3}+ 3

^{3}+â€¦..+ K

^{3}=

To Prove P(K+1) is true using P(k)

P(K + 1) = 1

^{3}+ 2

^{3}+ 3

^{3}+ K

^{3}+ (K + 1)

^{3}= â€¦â€¦â€¦.(1)

L.H.S of (1)

1

^{3}+ 2

^{3}+ 3

^{3}+ K

^{3}+ (K + 1)

^{3}= + (K +1)

^{3 }= + (K +1)

^{3 }=

^{ }=

^{ }=

^{ }=

=

R.H.S of ( 1).

Therefore P(K + 1) is proved

Hence by the principle of 4 mathematical induction P(n) is true for all the Values of n,

n N

Hence proved.

# Question-3

**1+.**

**Solution:**

Let P(n) = 1+.

P(1) is true

Let P (k) be true.

P(k) = 1+

To Prove P(K+1) is true using P(k) Lies

1+

= â€¦â€¦â€¦ (1)

L.H.S of (1)

=

Taking L.c.M

=

=

= =

= R.H.S of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-4

**Prove the following by the principle of mathematical induction for**

every natural number n.1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + n(n + 1)(n + 2) = n(n + 1)(n +2)(n +3)/4.

every natural number n.1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + n(n + 1)(n + 2) = n(n + 1)(n +2)(n +3)/4.

**Solution:**

Let P(n) : 1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + n(n + 1)(n + 2) = n(n + 1)(n +2)(n + 3)/4

P(1) : 1.2.3 = 1(1 + 1)(1 +2)(1 + 3)/4 = 1.2.3.4/4 = 6 which is true. Thus P(n) is true for n = 1.

Assume P(k) : 1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + k(k + 1)(k + 2) = k(k + 1)(k +2)(k + 3)/4

To prove: P(k + 1) : 1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + (k + 1)(k + 2)(k + 3) = (1/4)(k + 1)(k + 2)(k + 3)(k + 4)

L.H.S = 1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + (k + 1)(k + 2)(k + 3)

= 1.2.3 + 2.3.4 + 3.4.5 + â€¦.. + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

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

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

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

= R.H.S

Thus P(k + 1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for all natural numbers.

# Question-5

**1.3 +2.3**

^{2}+ 3.3^{3}+â€¦.+n.3^{n}=**Solution:**

Let P(n) = 1.3 +2.3

^{2}+ 3.3

^{3}+â€¦.+n.3

^{n}=

P(1) is true.

Let P(k) be true

P(k) = 1.3 +2.3

^{2}+ 3.3

^{3}+â€¦.+3.3

^{k}=

To prove P( k +1) is true using P(k)

P(k +1) = 1.3 +2.3

^{2}+ 3.3

^{3}+â€¦.+3.3

^{k}+ 3.3

^{k+1}

**=**

=

=â€¦â€¦â€¦â€¦.. (1)

=

=

L.H.s of ( 1)

1.3 + 2.3

^{2}+ 3.3

^{3 }+â€¦â€¦â€¦k.3

^{k}+ (k+1) 3

^{K+1}.

** = ** +(k+1) 3^{k+1}.

** =
=
=
=
= **

= R.H.S of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-6

**1.2 + 2.3 + 3.4 +â€¦. + n(n+1) =**

**Solution:**

P(1) is true

Let P(k) be true

1.2 + 2.3 + 3.4 +â€¦. + k(k+1) =

To prove P(k+1) to be true using the result of P(k)

1.2 + 2.3 + 3.4 +â€¦. + k(k+1) +(k+1)(k+2)

= â€¦â€¦â€¦(1)

L.H.S of (1)

1.2 + 2.3 + 3.4 +â€¦. + k(k+1) +(k+1)(k+2)

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

=

=

Which is the R.H.S of (1).

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-7

**1.3 + 3.5 + 5.7 +â€¦ + (2n â€“1) (2n + 1) =**

**Solution:**

P(1) is true

Let P (k) be true

1.3 + 3.5 + 5.7 +â€¦ + (2k â€“1) (2k + 1) =

To prove p(k+1) is true using P(k) lies

1.3 + 3.5 + 5.7 +â€¦ + (2k â€“1) (2k + 1)+[2(k+1)-1] [2(k+1)+1]

=

=

=

= â€¦â€¦â€¦.(1)

=

=

L. H.S of (1)

1.3 + 3.5 + 5.7 +â€¦ + (2k â€“1) (2k + 1)+[2(k+1)-1] [2(k+1)+1]

=

=

=

=

=

Which is the R. H.s of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-8

**1.2 +2.2**

^{2}+ 3.2^{2}+â€¦. +n.2^{n}= (n â€“ 1) 2^{n+1}+ 2.**Solution:**

P(1_ is trueAssuming P(k) is true.

1.2 +2.2

^{2}+ 3.2

^{2}+â€¦. +k.2

^{k}= (k â€“ 1) 2

^{k+1}+ 2

To prove P(k+1) is true using P(k) lies

1.2 +2.2

^{2}+ 3.2

^{2}+â€¦. +k.2

^{k}= (k + 1) 2

^{k+1}= (k)2

^{k+1+1}+ 2

= k.2

^{k+2}+ 2 â€¦â€¦â€¦â€¦â€¦(1)

L.H.S of (1)

1.2 + 2.2

^{2}+ 3.2

^{2}+â€¦â€¦â€¦..k.2

^{k}+ (k+1)2

^{(k+1)}(k-1)2

^{k+1}+ 2 + (k +1) 2

^{k+1 }= 2

^{k+1}(k+1+k-1)+2

= 2

^{k+1}2k + 2

= 2

^{k+2}.k + 2

= k.2

^{k+2}+ 2

Which is the R.H.s of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-9

**Solution:**

Let P(n) =

P(1) is true

Let us assume P(k) to be

To Prove P(k+1) is true using P(k) lies

â€¦â€¦â€¦â€¦.(1)

Taking L. H.S of (1)

= 1- = 1-

which is the R.H.s of (1)

P(K+1) is true. By the Principle of mathematical induction, P(n) is true for all values of n where n NHence proved.

# Question-10

**Solution:**

Let p(n) =

P(1) is true.

Let P(k) be true lies

P(k) =

To prove P(K+1) is true using P(k)

= â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)

L.H.S of (1)

=

=

=

=

=

=

=

= =

which is the R.H.S of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-11

**Solution:**

P(1) is trueLet us assume P(k) is true lies

To prove P(k+1) is true using P(k).â€¦â€¦â€¦â€¦â€¦..(1)

L.H.S of (1)

=

=

where is the R.H.S of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-12

**a + ar + ar**

^{2}+â€¦..+ar^{n-1}=**Solution:**

n (2) = a + ar = a(r+1)

R.H.S =

P(2) is true.

Let us assume p(k) is true lies

a + ar + ar

^{2}+â€¦..+ar

^{k-1}=

To Prove p(k+1) is true using P(k)

a + ar + ar

^{2}+â€¦..+ar

^{k-1}+ar

^{k+1-1}= â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

L.H.S of (1)

a + ar + ar

^{2}+â€¦..+ar

^{k-1}+ar

^{k}= a+r(a+ar+ar

^{2}+â€¦ar

^{k-1})

= a +

=

=

=

= =

Which is the R.H.S of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-13

**= (n+1)**

^{2}.**Solution:**

P(1) is true

Let us assume P(k) is true

= (k+1)

^{2}.

To prove P(k+1) is true using P(k)

P(k+1) =

= (k+2)

^{2}â€¦â€¦â€¦â€¦(1)

L.H.S of (1)

= (k+1)

^{2}

= (k+2)

^{2}

Which is the R.H.s of â€¦â€¦â€¦â€¦â€¦. (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-14

**= (n+1).**

**Solution:**

Let P(n) = = (n+1)

P(1) is true.

Assuming P(k) is true.

P(k) = = K+1

To prove P(k+1) is true.

P(k+1) = = K+2â€¦â€¦â€¦â€¦.(1)

L.H.s of (1)

= (k+1)1+

= = K+2

Which is the R.H.S of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-15

**1**

^{2}+ 3^{2}+5^{2}+ â€¦. +(2n â€“1)^{2}=**Solution:**

P(1) is True

Let us assume P(k) be true

1

^{2}+ 3

^{2}+5

^{2}+ â€¦. +(2k â€“1)

^{2}=

To prove P(k+1) is true using P(k)

P(k+1) =1

^{2}+ 3

^{2}+5

^{2}+ â€¦. +(2(k + 1)-1)

^{2}

=

= â€¦â€¦â€¦â€¦â€¦(1)

L.H.S of (1)

1

^{2}+ 3

^{2}+5

^{2}+ â€¦. +(2(k + 1)-1)

^{2}=

=

=

=

=

=

=

which is the R.H.S of â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-16

**Solution:**

P(1) is true

Let as assume P(k) is true

To prove P(k+1) is true using P(k).

P(k+1) =

**=**â€¦â€¦â€¦â€¦â€¦.(1)

L.H.S of (1)

=

=

=

=

=

= =

=

=

=

=

=

= =

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-17

**Solution:**

Let P(n) =

P(1) is true.

Let us assume P(k) is true.

To prove P(k+1) is true using P(k)

**â€¦.1**

=

= =

=

=

= =

=

Which is the R.H.s of (1)

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-18

**1 + 2+ 3 +â€¦. + n < ( 2n + 1)**

^{2 }**Solution:**

Let P(n) = 1 + 2+ 3 +â€¦. + n < ( 2n + 1)

^{2}

P(1) is true

Let us assume p(k) is true.

1 + 2+ 3 +â€¦. + k < ( 2k + 1)

^{2}

To prove P(k+1) is true using P(k)

P(k+1) = 1 + 2+ 3 +â€¦. + k +k+1< ( 2(k + 1)+1)

^{2 }= 1+ (1 + 2+ 3 +â€¦. + k) +k < ( 2k + 3)

^{2}â€¦â€¦.. 1

L .H.S 1+ ( 2k + 1)

^{2}+ k

=

^{ }= ( 4k

^{2}+ 1 +4k + 8k + 8)

^{ }= (4k

^{2}+ 12k +9)

^{ }= (2k + 3)

^{2}

Which is the R.H.S ofâ€¦â€¦. 1

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-19

**n(n+1) (n+5) ia a multiple of 3.**

**Solution:**

Let P(n) = n(n+1) (n+3)

P(1) is a multiple of 3

Let P(k) be a multiple of 3

(i.e) P(k) = K(k+1) (k+5) = 3mâ€¦â€¦â€¦â€¦..(A)

To prove P(k+1) is a multiple of 3 using resultâ€¦â€¦.(A)

P(k+1) = {(k+1)(k+2)}(k+6)

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

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

= 3m +

= 3m (1+ )

which is divisible by 3

hence the result.

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-20

**To Prove 10**

^{2n-1}+ 1 is divisible by 11.**Solution:**

10

^{2n-1}+ 1 is divisible by 11

Let P(n) = 10

^{2n-1}+ 1

P(1) divisible by 11

Let us assume P(k) = 10

^{2n-1}+ 1 is divisible by 11

P(k) = 10

^{2k-1}+ 1 = 11M â€¦â€¦â€¦â€¦â€¦â€¦â€¦( A)

To Prove P(k+1) is divisible by 11 using the result of A

P(k+1)

^{2}= 10

^{2(k+1)-1}+ 1

= 10

^{2k+2-1}+ 1

= 10

^{2k-1}. 10

^{2}+ 1

= (11M â€“ 1) 10

^{2}+ 1

= (10

^{2})11M â€“ 100 + 1

= (100)11M â€“ 99

= 11(100M â€“ 9)

Which is divisible by 11.

hence the result.

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

# Question-21

**x**

^{2n}â€“ y^{2n}is divisible by x + y.**Solution:**

Let P(n) = x

^{2n}â€“ y

^{2n}

P(1) = x

^{2}â€“ y

^{2 }= (x + y) (x-y) which is divisible by (x+y)

P(1) is true

Let us assume P(k) is true

P(k) = x

^{2k}-y

^{2k}= m(x+y)â€¦â€¦â€¦â€¦â€¦. A

TO Prove P(k+1) is divisible by (x+y) using the resultsof A

P(k +1) = x

^{2(k+1)}â€“ y

^{2(k+1)}

P(k+1) = x^{2(k+1)} â€“ y^{2(k+1)
}= x^{2k} . x^{2} â€“ y^{2k} . y^{2
}= x^{2k} . x^{2} â€“ [ x^{2k} â€“ m(x+y)]y^{2} From result A

^{ }= x^{2k} . x^{2} â€“ y^{2}.x^{2} + my^{2} (x+y)

^{ }= x^{2k} (x^{2} â€“ y^{2}) + my^{2} (x+y)

y^{2k } = x^{2k } - m(n+y)

= (x+y) { x^{2k} (x-y) + my^{2}}

P(k+1) is divisible by (x+y)** **

hence the result.

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of nwhere n N

Hence proved

# Question-22

**3**

^{2n+2}â€“ 8n â€“ 9 is divisible by 8.**Solution:**

Let P(n) = 3

^{2n+2}â€“ 8n â€“ 9

P(1) is divisible by 8

Let us assume P(k) is divisible by 8.

P(k) = 3

^{2k+2}â€“ 8k â€“ 9 = 8Mâ€¦â€¦â€¦â€¦.. (A)

To prove P(k+1) is divisible by 8 using the result of (A)

P(k+1) = 3

^{2(k+1)+2}â€“ 8(k+1) â€“ 9

= 3

^{2k+2+2}â€“ 8k â€“ 8 â€“9

= 3

^{2k+2}. 3

^{2}â€“ 8k- 9 â€“8

= (8M + 8K + 9) 9 â€“ 8k â€“ 9 â€“ 8

= 72 M + 72 K + 81 â€“ 8K â€“ 9 â€“ 8

= 72 M + 64 k +64

= 8(9M + 8K + 8)

P( K + 1) is divisible by 8.

hence the result.

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of nwhere n N

Hence proved

# Question-23

**41**

^{n}â€“ 14^{n}is a multiple of 27.**Solution:**

Let P(n) = 41

^{n}â€“ 14

^{n }

P(1) is a multiple of 27

Let us assume P(k) is a multiple of 27

P(k) = 41

^{k}â€“ 14

^{k}= 27Mâ€¦â€¦â€¦..(A)

To Prove P(k+1) is divisible by 27 using the result of ( A )

P(k +1) = 41

^{k+1}- 14

^{k+1 }= 41

^{k}.41 â€“ 14

^{k}. 14

^{ }= (27M + 14

^{k}) 41 â€“ 14

^{k}. 14

^{ }= 27M 41 +14

^{k}. 41 â€“ 14

^{k}. 14

^{ }= 27M 41 + 14

^{k}(41 â€“ 14)

^{ }= 27M 41 + 14

^{k}(27)

^{ }= 27(41M + 14

^{k})

P(k+1) is a multiple of 27.

hence the result.

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved

# Question-24

**(2n + 7) < (n+3)**

^{2}.**Solution:**

Let P(n) = (2n + 7) < (n+3)

^{2}

P(1) is true

Let P(k) = (2k+7) < (k+3)

^{2}

To Prove P(k+1) = (2(k+1)+7) < (k + 4)

^{2 }= [2k+9] < (k+4)

^{2 }= [(k+3)

^{2}â€“ 9] < (k + 4)

^{2 }= [( k+3+3) (k+3-3)] < ( k+4)

^{2 }= [(k+6)(k) < (k+4)

^{2 }= (k

^{2}+ 6k) < (k

^{2}+ 8k + 16)

The given expression is true for P(k+1)