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Question-1

1+3+32 +… +3n-1 = .

Solution:
Let P(n) = 1+3+32 +… +3n-1
P(1) is true
Assuming P(k) is true
1+3+32 +… +3k-1 =
To prove P(k +1) is true using the result of P(k)
1+3+32 +… +3k-1 +3k = ……………(1)
proof: Taking L.H.S of (1)
1+3+32 +… +3k-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

13 + 23 + 33 + …. +n3 = .

Solution:
Let P(n) = 13 + 23 + 33 + …. +n3 =
P(1) is true
Assume P(k) is true.
13 + 23 + 33 +…..+ K3 =
To Prove P(K+1) is true using P(k)
P(K + 1) = 13 + 23 + 33 + K3 + (K + 1)3 = ……….(1)
L.H.S of (1)
13 + 23 + 33 + K3 + (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+
                                                                    = ……… (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 + ….. + 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 + ….. + 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 + ….. + 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 + ….. + (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 + ….. + (k + 1)(k + 2)(k + 3)
        = 1.2.3 + 2.3.4 + 3.4.5 + ….. + 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 +2.32 + 3.33 +….+n.3n =

Solution:
Let P(n) = 1.3 +2.32 + 3.33 +….+n.3n =
P(1) is true.
Let P(k) be true
P(k) = 1.3 +2.32 + 3.33 +….+3.3k =
To prove P( k +1) is true using P(k)
P(k +1) = 1.3 +2.32 + 3.33 +….+3.3k + 3.3k+1

              =
              =
              =
………….. (1)
L.H.s of ( 1)
1.3 + 2.32 + 3.33 +………k.3k + (k+1) 3K+1.

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

Solution:
P(1) is true
Let P(k) be true
1.2 + 2.3 + 3.4 +…. + k(k+1) =
To prove P(k+1) to be true using the result of P(k)
1.2 + 2.3 + 3.4 +…. + k(k+1) +(k+1)(k+2)
                                          = ………(1)
L.H.S of (1)
1.2 + 2.3 + 3.4 +…. + 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 +… + (2n –1) (2n + 1) =

Solution:
P(1) is true
Let P (k) be true
1.3 + 3.5 + 5.7 +… + (2k –1) (2k + 1) =
To prove p(k+1) is true using P(k) lies
1.3 + 3.5 + 5.7 +… + (2k –1) (2k + 1)+[2(k+1)-1] [2(k+1)+1]
=
=
=
= ……….(1)
=
=
L. H.S of (1)
1.3 + 3.5 + 5.7 +… + (2k –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 +2.22 + 3.22 +…. +n.2n = (n – 1) 2n+1 + 2.

Solution:
P(1_ is trueAssuming P(k) is true.
1.2 +2.22 + 3.22 +…. +k.2k = (k – 1) 2 k+1 + 2
To prove P(k+1) is true using P(k) lies
1.2 +2.22 + 3.22 +…. +k.2k = (k + 1) 2 k+1 = (k)2 k+1+1 + 2
                                       = k.2k+2 + 2 ……………(1)
L.H.S of (1)
1.2 + 2.22 + 3.22 +………..k.2k + (k+1)2(k+1)(k-1)2k+1 + 2 + (k +1) 2k+1
                                              
= 2k+1 (k+1+k-1)+2
                                       = 2k+1 2k + 2
                                       = 2k+2.k + 2
                                       = k.2k+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
………….(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)
      = ………………….(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).……………..(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 + ar2 +…..+arn-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 + ar2 +…..+ark-1 =
To Prove p(k+1) is true using P(k)
a + ar + ar2 +…..+ark-1+ark+1-1 = …………………(1)
L.H.S of (1)
a + ar + ar2 +…..+ark-1+ark = a+r(a+ar+ar2+…ark-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…………(1)
L.H.S of (1)

= (k+1)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-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+2………….(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

12 + 32 +52 + …. +(2n –1)2 =

Solution:
P(1) is True
Let us assume P(k) be true
12 + 32 +52 + …. +(2k –1)2 =
To prove P(k+1) is true using P(k)
P(k+1) =12 + 32 +52 + …. +(2(k + 1)-1)2  
          =
          = ……………(1)
L.H.S of (1)
12 + 32 +52 + …. +(2(k + 1)-1)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-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) =

            = …………….(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)

….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 +…. + n < ( 2n + 1)2

Solution:
Let P(n) = 1 + 2+ 3 +…. + n < ( 2n + 1)2
P(1) is true
Let us assume p(k) is true.
1 + 2+ 3 +…. + k < ( 2k + 1)2
To prove P(k+1) is true using P(k)
P(k+1) = 1 + 2+ 3 +…. + k +k+1< ( 2(k + 1)+1)2
           
= 1+ (1 + 2+ 3 +…. + k) +k < ( 2k + 3)2…….. 1
L .H.S 1+ ( 2k + 1)2 + k
          =
            = ( 4k2 + 1 +4k + 8k + 8)
            = (4k2 + 12k +9)
            = (2k + 3)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-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) = 3m…………..(A)
To prove P(k+1) is a multiple of 3 using result…….(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 102n-1 + 1 is divisible by 11.

Solution:
102n-1 + 1 is divisible by 11
Let P(n) = 102n-1 + 1
P(1) divisible by 11
Let us assume P(k) = 102n-1 + 1 is divisible by 11
P(k) = 102k-1 + 1 = 11M …………………( A)
To Prove P(k+1) is divisible by 11 using the result of A
P(k+1)2 = 102(k+1)-1 + 1
            = 102k+2-1 + 1
            = 102k-1 . 102 + 1
            = (11M – 1) 102 + 1
            = (102)11M – 100 + 1
            = (100)11M – 99
            = 11(100M – 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 – y2n is divisible by x + y.

Solution:
Let P(n) = x 2n – y2n
P(1) = x2 – y2
       
= (x + y) (x-y) which is divisible by (x+y)
P(1) is true
Let us assume P(k) is true
P(k) = x2k-y2k = m(x+y)……………. A
TO Prove P(k+1) is divisible by (x+y) using the resultsof A
P(k +1) = x2(k+1) – y2(k+1)

P(k+1) = x2(k+1) – y2(k+1)
           
= x2k . x2 – y2k . y2
           
= x2k . x2 – [ x2k – m(x+y)]y2 From result A
            = x2k . x2 – y2.x2 + my2 (x+y)
            = x2k (x2 – y2) + my2 (x+y)
y2k = x2k - m(n+y)
     = (x+y) { x2k (x-y) + my2}
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

32n+2 – 8n – 9 is divisible by 8.

Solution:
Let P(n) = 32n+2 – 8n – 9
P(1) is divisible by 8
Let us assume P(k) is divisible by 8.
P(k) = 32k+2 – 8k – 9 = 8M………….. (A)
To prove P(k+1) is divisible by 8 using the result of (A)
P(k+1) = 32(k+1)+2 – 8(k+1) – 9
          = 32k+2+2 – 8k – 8 –9
          = 32k+2 . 32 – 8k- 9 –8
          = (8M + 8K + 9) 9 – 8k – 9 – 8
          = 72 M + 72 K + 81 – 8K – 9 – 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

41n – 14n is a multiple of 27.

Solution:
Let P(n) = 41n – 14n
P(1) is a multiple of 27
Let us assume P(k) is a multiple of 27
P(k) = 41k – 14k = 27M………..(A)
To Prove P(k+1) is divisible by 27 using the result of ( A )
P(k +1) = 41k+1 - 14k+1
            
= 41k .41 – 14k . 14
             = (27M + 14k) 41 – 14k . 14
             = 27M 41 +14k . 41 – 14k . 14
             = 27M 41 + 14k (41 – 14)
             = 27M 41 + 14k (27)
             = 27(41M + 14k)
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 – 9] < (k + 4)2
                           
= [( k+3+3) (k+3-3)] < ( k+4)2
                           
= [(k+6)(k) < (k+4)2
                           
= (k2 + 6k) < (k2 + 8k + 16)
The given expression is true for P(k+1)




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