TOPICS
Exercise - 4.1

Question-1 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
  Let the given statement be P(n), i.e., P(n): 1 + 3 + 32 + ...... + 3n-1 = (3n - 1)/2
  For n = 1, we have, P(1): 1 =	(31 - 1)/2 = 1, which is true. 
        
  Let P(k) be true for some positive integer k, i.e., P(n): 1 + 3 + 32 + ...... + 3k-1 = (3k - 1)/2

  We shall now prove that P(k + 1) is true.
  Now, 1 + 3 + 32 + ....... + 3k-1 + 3k+1-1 
= (1 + 3 + 32 +.... + 3k-1) + 3k
= (3k - 1)/2 + 3k
= {(3k - 1) + 2.3k}/2
= {(1 + 2).3k - 1}/2
= (3.3k - 1)/2
= (3k+1 - 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 i.e., n.
   

Question-2 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
       PMI
   

Question-3 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
    

Question-4 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-5 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-6 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
    

Question-7 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-8 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
  Let the given statement be P(n), i.e., P(n): 1.2 + 2.22 + 3.22 + ..... + n.2n = (n – 1)2n+1 + 2 For n = 1, 
  we have, P(1): 1.2 = 2 = (1 – 1).21+1 + 2 = 0 + 2 = 2, which is true. 

  Let P(k) be true for some positive integer k, i.e., 1.2 + 2.22 + 3.22 + ..... + k.2k = (k – 1)2k+1 + 2 .......(i) 

  We shall now prove that P(k + 1) is true.
  Now, {1.2 + 2.22 + 3.22 + ..... + k.2k} + (k+1)2k+1
= (k - 1).2k+1 + 2 + (k + 1).2k+1
= 2k+1{(k - 1) + (k + 1)} + 2
= 2k+1. 2k + 2
= k.2(k+1)+1 + 2
= {(k+1) - 1}.2(k+1)+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 i.e., n.
   

Question-9 :-  Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
  Let the given statement be P(n), i.e., P(n): 1/2 + 1/4 + 1/8 + ......+ 1/2n = 1 - 1/2n
  For n = 1, we have P(1): 1/2 = 1 - 1/21 = 1/2, which is true.

  Let P(k) be true for some positive integer k, i.e., 1/2 + 1/4 + 1/8 + ......+ 1/2k = 1 - 1/2k .......(i)

  We shall now prove that P(k + 1) is true.
  Now, (1/2 + 1/4 + 1/8 + ......+ 1/2k) + 1/2k+1
= (1 - 1/2k) + 1/2k+1         [using i]
= 1 - 1/2k + 1/2.2k
= 1 - 1/2k(1 - 1/2)
= 1 - 1/2k. 1/2
= 1 - 1/2k+1

  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 i.e., n.
    

Question-10 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-11 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-12 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-13 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-14 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
  Let the given statement be P(n), i.e., P(n) : (1 + 1/1).(1 + 1/2).(1 + 1/3)......(1 + 1/n) = (n + 1)
  For n = 1, we have (1 + 1/1) = (1 + 1) = 2, which is true.

  Let P(k) be true for some positive integer k, i.e., P(k): (1 + 1/1).(1 + 1/2).(1 + 1/3)  ......(1 + 1/k) = (k + 1)

  We shall now prove that P(k + 1) is true. 
  Now, {(1 + 1/1).(1 + 1/2).(1 + 1/3)...... (1 + 1/k)} + {1 + 1/(k+1)}
= (k + 1).{1 + 1/(k+1)}
= (k + 1).{(k+1) + 1}/(k+1)
= (k+1) + 1

  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 i.e., n.
   

Question-15 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-16 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-17 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
        PMI
   

Question-18 :- Prove the following by using the principle of mathematical induction for all n ∈ N: PMI

Solution :-
  Let the given statement be P(n), i.e., P(n) : 1 + 2 + 3 + .......+ n < 1/8.(2n + 1)2
  It can be noted that P(n) is true for n = 1 since 1 < 1/8.(2.1 + 1)2 = 9/8. 

  Let P(k) be true for some positive integer k, i.e., 1 + 2 + 3 + .....+ k < 1/8.(2k + 1)2  ......(1)

  We shall now prove that P(k + 1) is true whenever P(k) is true.
  Now, (1 + 2 + 3 +  ... + k) + (k + 1) < 1/8.(2k + 1)2 + (k + 1)  [By using 1]
< 1/8{(2k + 1)2 + 8(k + 1)}
< 1/8{(4k2 + 1 + 4k + 8k + 8)}
< 1/8{(4k2 + 12k + 9)}
< 1/8.(2k + 3)2
< 1/8{2(k+1) + 1}2

  Hence, (1 + 2 + 3 +  ... + k) + (k + 1) < 1/8.(2k + 1)2 + (k + 1)
  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 i.e., n.
   

Question-19 :-  Prove the following by using the principle of mathematical induction for all n ∈ N:
n(n + 1)(n + 5) is a multiple of 3.

Solution :-
  Let the given statement be P(n), i.e., 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.

  Let P(k) be true for some positive integer k, i.e., k (k + 1) (k + 5) is a multiple of 3.
  Therefore, 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.
  Now, (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) + (k + 1)(k + 2)}
= 3m + (k + 1){2(k + 5) + (k + 2)}
= 3m + (k + 1){2k + 10 + k + 2}
= 3m + (k + 1){3k + 12}
= 3m + 3(k + 1)(k + 4)
= 3{m + (k + 1)(k + 4)}
= 3q      where q = m + (k + 1)(k + 4) is a natural number.

  Therefore, (k + 1){(k + 1) + 1}{(k + 1) + 5} is divisible by 3.
  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 i.e., n.
   

Question-20 :-  Prove the following by using the principle of mathematical induction for all n ∈ N:
102n-1 + 1 is divisible by 11.

Solution :-
  Let the given statement be P(n), i.e., P(n): 102n-1 + 1 is divisible by 11.
  It can be observed that P(n) is true for n = 1 since P(1) = 102.1-1 + 1 = 11, which is divisible by 11.

  Let P(k) be true for some positive integer k, i.e., 102k-1 + 1 is divisible by 11.

  Therefore, 102k-1 + 1 = 11m, where m ∈ N ........ (1)
  We shall now prove that P(k + 1) is true whenever P(k) is true.
  Now, 102(k+1)-1 + 1
= 102k+1 + 1
= 102k. 102 + 1
= 102(102k + 1 - 1) + 1
= 102(102k + 1) - 102 + 1
= 102. 11m - 100 + 1
= 100 . 11m - 99
= 11(100m - 9)
= 11r,    where r = 100m - 9 is some natural number.
  
  Therefore, 102(k+1)-1 + 1 is divisible by 11.
  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 i.e., n.
   

Question-21 :- Prove the following by using the principle of mathematical induction for all n ∈ N:
x2n – y2n is divisible by x + y.

Solution :-
  Let the given statement be P(n), i.e., P(n): x2n – y2n is divisible by x + y.
  It can be observed that P(n) is true for n = 1.
  This is so because x2 x 1 – y2 x 1 = x2 – y2 = (x + y) (x – y) is divisible by (x + y). 

  Let P(k) be true for some positive integer k, i.e.,
  x2k – y2k is divisible by x + y.
  Therefore, x2k – y2k = m (x + y), where m ∈ N ........ (1)

  We shall now prove that P(k + 1) is true whenever P(k) is true.
  Now, x2(k+1) - y2(k+1)
= x2k x2 - y2k y2
= x2(x2k - y2k + y2k) - y2k y2
= x2(m(x + y) + y2k) - y2k y2
= x2 . m(x + y) + y2k . x2 - y2k y2
= x2 . m(x + y) + y2k. (x2 - y2)
= x2 . m(x + y) + y2k. (x - y)(x + y)
= (x + y)[x2 . m + y2k. (x - y)]      which is a factor of (x + y)

  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 i.e., n.
   

Question-22 :-  Prove the following by using the principle of mathematical induction for all n ∈ N:
32n+2 – 8n – 9 is divisible by 8.

Solution :-
  Let the given statement be P(n), i.e., P(n): 32n+2 – 8n – 9 is divisible by 8.
  It can be observed that P(n) is true for n = 1 since 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8.

  Let P(k) be true for some positive integer k, i.e., 32k+2 – 8k – 9 is divisible by 8.
  Therefore, 32k+2 – 8k – 9 = 8m; where m ∈ N ........(1)

  We shall now prove that P(k + 1) is true whenever P(k) is true.
  Now, 32(k+1)+2 – 8(k+1) – 9
= 32k+2. 32 – 8k - 8 – 9
= 32(32k+2 + 9 + 8k - 9 - 8k) – 8k - 17
= 32(32k+2 - 9 - 8k) + 32(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

  Therefore, 32(k+1)+2 – 8(k+1) – 9 is divisible by 8.
  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 i.e., n.
   

Question-23 :-  Prove the following by using the principle of mathematical induction for all n ∈ N:
41n – 14n is a multiple of 27.

Solution :-
  Let the given statement be P(n), i.e., P(n): 41n – 14n is a multiple of 27.
  It can be observed that P(n) is true for n = 1 since 411 - 141 = 27, which is a multiple of 27.

  Let P(k) be true for some positive integer k, i.e., 41k – 14k is a multiple of 27
  Therefore, 41k – 14k = 27m, where m ∈ N ....... (1)
  We shall now prove that P(k + 1) is true whenever P(k) is true.
  Now, 41k+1 - 14k+1
= 41k. 41 - 14k. 14
= 41.(41k + 14k - 14k) - 14k. 14
= 41.(41k - 14k) + 41 . 14k - 14k. 14
= 41.27m + 14k(41 - 14)
= 41.27m + 14k. 27
= 27(41m + 14k)
= 27 x r    where r = (41m + 14k) is a natural number

  Therefore, 41k+1 – 14k+1 is a multiple of 27
  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 i.e., n.
   

Question-24 :- Prove the following by using the principle of mathematical induction for all n ∈ N:
(2n + 7) < (n + 3)2

Solution :-
  Let the given statement be P(n), i.e., 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)2 = 16, which is true.

  Let P(k) be true for some positive integer k, i.e., (2k + 7) < (k + 3)2 + 2 .....(1)

  We shall now prove that P(k + 1) is true whenever P(k) is true. 
  Now, 2(k + 1) + 7 = (2k + 7) + 2
  Therefore, 2(k + 1) + 7 = (2k + 7) + 2 < (k + 3)2
  2(k + 1) + 7 < k2 + 6k + 9 + 2
  2(k + 1) + 7 < k2 + 6k + 11
  Now, k2 + 6k + 11 < k2 + 8k + 16
  Therefore, 2(k + 1) + 7 < (k + 4)2
  2(k + 1) + 7 < {(k + 1) + 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 i.e., n.
   
CLASSES

Connect with us:

Copyright © 2015-17 by a1classes. All Rights Reserved.

www.000webhost.com