Chapter -2 Mathematical Induction
INTRODUCTION: Method of mathematical induction is a very powerful method for proving certain
results involving natural numbers. It is a coordination of two concepts, numerical supports (s) and
induction step. The principle of mathematical induction states that “if a statement P(n) is true for
n = 1 and whenever P(n) is true P(n + 1) is true, then the statement P(n) is true for all natural
numbers n”. Alternatively if P(1) be true and P(n) 
P(n + 1), then, P(n) is true for all n.
Following results and instructions will be useful throughout this chapter on induction.
1.
The least n for which a result is true may not be 1 in all the problems. For instance, the result
2n > n2 is true for all n ≥ 5. Thus, the least n for which is the true is 5. In proving this inequality
by induction we verify it for n = 5.
2.
If a result is required for certain types of n only, then assuming P(n) we prove some other
statement and not P(n + 1) necessarily. For instance, “Suppose u1 = u2 = 1 and un = un – 1 + un – 2 if
n > 2, then prove that un is divisible by 5, if n is divisible by 5”. Indeed, we are required to show
the result for n = 5, 10, 15, ……… Thus assuming P(n), we will try to prove P(n + 5).
3.
If P(n) is true 
P(n + r) is true then the result will be true, for all n provided it is true for
n = 1, 2, 3, …………, r.
4.
If a result be true for n = 1 and n = 2 and if under the assumption that it is true for n = k – 2,
n = k – 1, we prove it for n = k, then the result is true for all natural numbers n. (Principle of finite
induction). Equivalently if P(1), P(2) be true and whenever P(k – 2), P(k – 1) are true, P(k) is also
true, then the result is true for all n.
5.
The above principle is equivalent to the following statement.
“ If P(1) be true and if under the assumption that it is true for all positive integers less than n, we
prove it for n, then the result is true for all n”.
6.
If P(n) 
P(n + 1) and P(1) is not true of all n (or for any n). For instance “If 4n + 7n is
divisible by 3, prove that 4n+1 + 7n+1 is also divisible by 3”. We have
4n 1  7 n 1  4.4n  4.7 n  4.7 n  7 n 1  4  4n  7 n   7 n  3 ………………………..(A)
Which shows that 4n 1  7 n 1 is divisible by 3. But P(1) is not true, since 11 is not divisible by 3.
Indeed P(n) is not true for all n. The relation (A) shows that if 4n  7 n is not divisible by 3, then
7.
4n 1  7 n 1 is not divisible by 3 either. Hence the reversed result follows from mathematical
induction.
If a result is true for few values (or even thousand values) then no conclusion can be drawn
regarding the truthness of result for all n. Following are some illustrations.
(i)
n2 – n + 41 is prime for n = 1, 2, ………, 40 but at n = 41 it is not a prime.
(ii)
8.
22  1 is prime for n = 1, 2, 3, 4 but it is not a prime for n = 5
2 p1  1 is not divisible by p2 for any prime p. The statement is true for all p which are
n
(iii)
less than one thousand. It was found that at p = 1093 the statement is false. i.e., 21092 – 1 is
divisible by 10932.
(iv)
“991n2 + 1 is not a perfect square”. The least n for which the result is false is
n = 12055735790331354447442536767.
If A > B is given and we want to prove that A > C, then to arrive at A > C, it is enough to show
that B ≥ C. (Remember that for the truthness of A > C, it is not necessary that B ≥ C).
1
9.
If A < B is given and we want to arrive at A < C, then to arrive at A < C, it is enough to show that
B ≤ C.
10.
For all n, 2  1 


n
1
  3.
n
Solved Examples
If f(n) = 6n+2 + 72n+1 where n is a natural number. Prove that f(n + 1) = 6f(n) + 43 . 72n+1 hence
show that f(n) is divisible by 43 for all natural numbers n.
Solution: We have
f(n) = 6n+2 + 72n+1
…….. (1)
n+3
2n+3
f(n + 1) = 6 + 7
…….. (2)
On multiplying (1) by 6, we get
6f(n) = 6n+3 + 6.72n+1
……..(3)
On subtracting (3) from (2), we get
f(n + 1) – 6f(n) = 72n+3 – 6.72n+1
= 72n+1 (72 – 6)

f(n + 1)
= 6f(n) + 43 . 72n+1
……..(A)
3
3
Now f(1)
=6 +7
= 559
Which is divisible by 43 and (A) shows that if f(n) is divisible by 43 then f(n + 1) is also
divisible by 43.

f(n) is also divisible by 43 for all n.
2.
State example 1 in the conventional style and give a conventional proof.
Solution: The conventional style of the problem 1 is Prove that 6n+2 + 72n+1 is divisible by 43 for all
natural numbers n.
Conventional Proof: The statement is true for n = 1. Since 63 + 73 = 559 which is divisible by 43.
Let the statement be true for n = k, then 6k+2 + 72k+1 is divisible by 43.
To complete the induction we must show the validity of the statement for n = k + 1. i.e., we must
show that 6k+3 + 72k+3 is divisible by 43. Indeed
6k+3 + 72k+3
= 6.6k+2 + 6.72k+1 – 6.72k+1 + 72k+3
= 6(6k+2 + 72k+1) + 72k+1(-6 + 72)
= 6(6k+2 + 72k+1) – 43.72k+1

6k+3 + 72k+3 is divisible by 43. Which proves the assertion
3.
If n is even prove that 52n -1 is divisible by 13.
Solution: Let f(n) = 52n – 1. Then f(2) = 54 – 1 = 624 which is divisible by 13.
Now if n is even then the next even is n + 2, we try to establish a relation between f(n) and
f(n + 2). Indeed f(n + 2) = 52n+4 – 1. And 54 f(n) = 52n+4 – 54.
On subtracting, we get
f(n + 2) – 54 f(n) = -1 + 54

f(n + 2) = 54 f(n) + 624

f(n + 2) is divisible by 13 whenever f(n) is divisible by 13
Therefore with the help of f(2), it follows that f(2), f(4), f(6), ………, etc. are divisible by 13,
equivalently 52n – 1 is divisible by 13 if n is even.
4.
If f(n) = 32n + 3n + 1 then show that f(n + 3) = 36 f(n) – 702 . 3n – 728. Hence show that f(n) is
1.
2
(a)
divisible by 13 if n = 1, 4, 7, …………..
(b)
divisible by 13 if n = 2, 5, 8, …………..
(c)
not divisible by 13 if n = 3, 6, 9, ……………
Solution: As earlier f(n + 3) = 32n+6 + 3n+3 + 1
36 f(n) = 32n+6 + 36 . 3n + 36

f(n + 3) – 36 f(n) = 3n(33 – 36) + 1 – 36

f(n + 3) = 36 f(n) – 702 . 3n – 728
…………(A)
Now (a) follows since f(1) = 13. Which is divisible by 13 and (A) shows that if f(n) is divisible by
13 then f(n + 3) is also divisible by 13. Similarly (b) follows since f(2) = 91.
To prove (c) we note that f(3) = 36 + 33 + 1 = 757
Which is not divisible by 13 and (A) shows that if f(n) is not divisible by 13, then f(n + 3) is also
not divisible by 13.
5.
Prove that the number 22  1 ends in the digit 7 for all n > 1.
n
Solution: For n = 2 the given expression (say f(n)) = 22  1 = 17. Which ends in the digit 7.
Let f(n) end in the digit 7 for some fixed value of n then to complete induction we observe that
n
f(n + 1) = 22  1
n1
 
= 22
n
2
1
(note)
= (f(n) – 1)2 + 1
Now f(n) – 1 ends in the digit 6

f(n) – 1 ends in the digit 6

(f(n) – 1)2 ends in the digit 6

(f(n) – 1)2 + 1 ends in the digit 6

f(n + 1) ends in the digit 7.
6.
Prove by induction that 12 + 32 + 52 + …………… + (2n – 1)2 = n/3 (4n2 – 1)
Solution: Denoting LHS by f(n) and RHS by g(n) we note that f(n) is sum of squares of first n odd
numbers and we have to prove that f(n) = g(n) for all n.
f(1)
= 12, g(1) = (1/3)(4.12 – 1)

f(1)
= g(1)
Suppose f(n) = g(n) for some fixed n, then
f(n + 1)
= sum of squares of first n + 1 odd numbers
= 12 + 32 + 52 + ………….. + (2n – 1)2 + (2n + 1)2
n
2
4n 2  1   2n  1

3
2n  1
2n  1
2n 2  5n  3
=
 n  2n  1  3  2n  1  

3
3
2n  1
2n  1
2n 2  2n  3n  3 
=
 2n  3 n  1

3
3
n 1
2
4  n  1  1
 g  n  1
=
3
= g  n    2n  1


2

Whence the result follows by mathematical induction.
7.
Prove that 12 – 32 + 52 – 72 + …………… - (4n – 1)2 = -8n2
Solution: As earlier denoting LHS by f(n) and RHS by g(n) we have (note that f(n) is alternating sum of
squares of consecutive odd numbers from 1 to 4n – 1)
3
f(1) = 12 – 32 = -8
and
g(1) = -8 . 12 = -8

f(1) = g(1)
Suppose f(n) = g(n) for some n, then
f(n + 1) = 12 – 32 + 52 – 72 + ………….. – (4n – 1)2 + (4n + 1)2 – (4n + 3)2
(Note carefully)
= g(n) + (4n + 1)2 – (4n + 3)2 = -8n2 + (4n + 1)2 – (4n + 3)2
= -8 (n + 1)2 = g(n + 1)

f(n + 1) = g(n + 1)
Which completes the induction.
8.
Prove by induction that
1 1 1
1
1
 2  2  ........  2  2  where n is a natural number greater
2
1 2 3
n
n
than 1.
Solution : The given inequality is true for n = 2 Since
1 1
1
 2  2
2
1 2
n
Let the given inequality be true for some fixed n, then
1 1 1
1
1
 2  2  ........  2  2 
2
1 2 3
n
n
………..(A)
To complete the induction we must show the result for n + 1 i.e., we must show that
1 1 1
1
1
1
 2  2  ........  2 
 2
2
2
1 2 3
n  n  1
n 1
Or we must show that
1 1 1
1
1
1
 2  2  ........  2  2 

2
1 2 3
n
n  1  n  12
In the light of induction hypothesis (A) it is now sufficient to show that
2
Or
1
1
1
 2

n
n  1  n  12
1
1
1


n n  1  n  12
Or
 n  1
2
 n  n  1  n Or
n2 + 2n + 1 ≥ n2 + 2n
Which is true for all n.

The inequality is true for n + 1 as well.
9.
2n !
Prove by induction
4  n !
n
2

1
.
3n  1
Solution: The inequality is true for n = 1, since 2/4 ≤ 1/2
Suppose for some fixed n
2n !
4n  n !
2

1
3n  1
………..(A)
To prove the inequality for n + 1 we must show that
 2n  2  !
2
4n 1   n  1 !

1
3n  4
Or, we must show that
4
 2n  2  2n  1 2n ! 
2
2
4  4n  n  1  n !
1
3n  4
Or, we must show that
2n !
4  n !
n
2

2n  2
 2n  1 3n  1
In the light of (A), it is now sufficient to show that
1
2n  2

3n  1  2n  1 3n  4
Or
(2n + 1)2(3n + 4) ≤ (2n + 2)2(3n + 1)
Which is true for all n

The inequality is true for n + 1 as well.
10.
19n ≤ 20n
Or
Prove by induction that 1. nC1  2. nC2  3.n C3  ..................n. nCn  n.2n 1 .
Solution : We have f 1  1. 1C1 ,
g 1  1.211
(Also f  2  1. 1C1  2. 2C2 , f  3  1. 3C1  2. 3C2  3. 3C3 , etc.)
As before assume f  n   g  n  for some n , then
f  n  1  1. n 1C1  2.n 1 C2  3.n 1 C3  ....................n.n 1 Cn   n  1
n 1
Cn 1
 1.  n C1  nC0   2.  n C2  nC1   3  n C3  nC2   .............  n  n Cn  nCn1    n  1 Cn
n
(On replacing
n 1
Cn 1  1 by n C n )
 1. nC1  2. nC2  3.n C3  .........  n.n Cn   1. nC0  2. nC1  3n C2  .........  n. nCn 1   n  1 nCn
 f  n    n C0  nC1  nC2  .........  nCn   1. nC1  2. nC2  3. nC3  .........  n nCn 
 f  n   2n  f  n 
 2n 2n1  2n
11.
 2 f  n   2n
 2n  n  1
 2g  n   2n
(
f  n   g  n  by hypothesis)
 g  n  1 .
If U1  6, U 2  28 and U n 1  6U n  4U n 1, prove that U n is an integer which is divisible by 2 n
Solution : We note that U1 , U 2 are integers if U n , U n1 are integers then U n 1 is also an integer.
Thus the result follows from Second principle of induction.
To prove the second part we note that U1 is divisible by 21 , and U 2 is divisible by 2 2 .
Let us assume that U n 1 is divisible by 2n1 i.e. U n 1  λ 2n 1 and U n is divisible by 2 n .
i.e U n  μ 2n
(Where λ, μ are integers)
Now we must show that U n 1 is divisible by 2n1. Indeed
Un1  6Un  4Un1  6 μ2n  4 λ2n1  3μ  λ  2n1  U n1 is divisible by 2n.
Another form : Show that sum of nth powers of roots of the equation x 2  6 x  4  0 is an integer
which is divisible by 2 n .
5
Solution :
If x1 and x2 be the roots then x1  x2  6, x1.x2  4.
And if U n denotes the sum of nth powers of roots then
U n  x1n  x2n ,
U1  6, U 2  x12  x22  28
On multiplying both sides by x1  x2, we get
  x1  x2   x1n  x2n 
 x1  x2 U n
 x1n 1  x2n 1  x1.x2n  x2 .x1n  x1n 1  x2n 1  x1.x2 ( x1n 1  x2n 1 )
 6U n

 6U n  4U n1
U n1
Now proceed as above to prove by induction.

Prove that integer just greater than 3  5
Yet another form :
Solution :



n
is divisible by 2n1
n
If U n be the integer just greater than 3  5 then

  3  5 
Since 0   3  5   1 and U is actually an integer which is
2 3  C 3  5   C 3  5   ......


Un  3  5
n
n
n
n
n
n 2
n
2
4
n 4
n
2
4
Now, let x1  3  5, x2  3  5.
U n  x1n  x2n
Then x1  x2  6, x1.x2  4, then
Now we have to proceed as above.
12.
If u1  u2  1, un  un1  un2
 n  2 . Show that
(i)
u12  u22  u32  ...........un2  unun 1
(iii)
n
n
1  1  5   1  5  

un 
 
 
5  2   2  


Solution : (i)
(ii)
2n 1 un  n is divisible by 5 .
Denoting LHS by f  n  and RHS by g  n  we note that
f 1  u12  1,

g 1  u1u2  1
f 1  g 1

 

Now f  n  1  f  n   u1  u22  .......un2  un21  u12  u22  .......  un2  un21
and g  n  1  g  n   un1un2  unun1
 un1 un2  un 
 un1  un  un1  un   un21
 f  n  1  f  n   g  n  1  g  n  for all n
Thus if f  n   g  n  for some n then  f  n  1  g  n  1
6
whence the result follows.
(ii)
Let f  n  = 2n 1U n  n
We easily prove that f 1 , f  2 are divisible by 5 since f 1  0, f  2  0
Let f  n  and f  n 1 be divisible by 5. Then
2n2U n1   n 1  5l
2n 1U n  n  5l '
and
where l and l ' are integers.
On multiplying the first relation by 2 and adding it into the second relation, we get
2n1 U n  U n1   3n  2  10l  5l '
On multiplying by 2, we get 2n U n 1  6n  4  20l  10l '
 2nUn1   n  1  20l  10l ' 5n  5  multiple of 5.
which show that f  n  1 is divisible by 5.
(iii)
2
2
1  1  5   1  5  
1  1  5   1  5  

We have U1 

 
   1.

  1 , U 2 
5  2   2  
5  2   2  


 The result is true for n  1, n  2
Suppose for some n,
where a 
Now U n
U n2 
1
1 n 1 n 1
a n 2  b n 2  , U n 1 

a  b 
5
5
1 5
1 5
,b
2
2
1 n 1 n 1
1 n2 n2
 U n 1  U n 2 
a b 

a  b 
5
5
1 n2
1 n2

a 1  a  
b 1  b 
5
5
2
Now
1 5 3  5  1 5 
2
1 a  1

 
  a
2
2
 2 
2
1 5 3  5 1 5 
2
1 b  1

 
  b
2
2
 2 
1 n n
 Un 
 a  b  which completes induction.
5
The sequence in the above question is the famous Fibonacci sequence.
13.
Each of the persons who ever lived on earth has shaken hands a certain number of times. Prove
that any time the number of those persons who have shaken hands an odd number of times is
even.
Solution: At any instant in our civilization there are two types of persons (including those who do not
exist at that instant)
7
Type A: Those persons who have shaken hands an odd number of times till that instant.
Type B: Those persons who have shaken hands an even number of times till that instant.
We have to show that at any instant the number of persons falling in type A category is even.
Suppose n hand shakes have already been taken place till that instant and number of persons in
Type A category is even then the (n + 1)th hand shakes can take place in three ways.
(1)
between any two persons of type A
(2)
between any two persons of type B
(3)
between one person of type A and one person of type B
If (1) be the case then after (n + 1) the hand shake the number of persons in type A
= Original number – 2 = Even – 2
= Even.
If (2) be the case then after (n + 1) the hand shake the number of persons in type (A)
= Original number + 2 = Even + 2
= Even.
And if (3) be the case then after (n + 1)th hand shake the number of persons in type (A) remains
same (since persons making (n + 1)th hand shake interchange).
Thus the number remains even in all cases.
Finally we note that the result is true for n = 1. Since the first hand shake took place between
two persons after which the number of persons in type A must have become Even.
Exercise
1.
Prove the following divisibility problems by mathematical induction. Unless the contrary is
specified prove it for all natural number n.
7 2 n  1 : 48
(i)
(iv)
(vii)
(x)
(xiii)
(xvi)
22n – 3n – 1 : 9
10n + 3.4n+2 + 5 : 9
2 .7n + 3.5n – 5 : 24
62n + 3n+2 + 3n : 11
11n+2 + 122n+1 : 133
52n + 1 : 13 if n is odd
(xix)
n(n  1)(n  5) is a multiple of 3
(xxi)
32n 2  8n  9 is divisible by 8. (xxii)
(ii)
(v)
(viii)
(xi)
(xiv)
(xvii)
(iii)
102n-1 + 1 : 11
4n + 15n – 1 : 9
(vi)
72n + 23n-3 3n-1 : 25
2n+1
3
+ 40n – 67 : 64
(ix)
52n+1 + 2n+4 + 2n+1 : 23
n3 + (n + 1)3 + (n + 2)3 : 9 (xii) n3 + 3n2 + 5n + 3 : 3
n3 – n : 24 if n is odd (xv)
22n+1 – 9n2 + 3n-2 : 54
33n+1 + 2n+1 : 5 (xviii) 4n+1 + 52n-1 : 21
(xx)
x 2 n  y 2 n is divisible by x+ y
41n  14n is multiple of 27
2.
3.
If n = 3k prove that 52n + 5n + 1 is not divisible by 31.
Show that 4.6n + 5n+1 leaves remainder 9 when divided by 20.
4.
Show that 8.7 n  4 n  2 is an odd multiple of 24 for all n.
5.
6.
Show that 32  1 is not divisible by 2n+3 for all n.
Show that n7 – 7n5 + 14n3 – 8n is divisible by 840 for all n.
7.
n5 n3 7 n
Show that the expression
is a positive integer for all n.
 
5 3 15
8.
Show that x x n 1  na n 1  a n  n  1 is divisible by (x – a)2 for all positive integers n > 1
9.
Prove that the length
n


n is constructable by ruler and compass.
8
n
2


nπ
nπ 
 i sin 
4
4 
10.
Prove that 1  i   2  cos
11.
Prove that the following inequalities by mathematical induction.
n
2n
(i)
Cn
1

n
4
2n  1
(iv)
n4 < 10n (n ≥ 2) (v)
(vi)
 n 1
n!  
 , n 1
 2 
(ii)
 2n  !  4 n n  1
 
 n ! n  1
1 + 2 + 3 + ………..+ n <
(iii)
n
2
n  n ! n  3
1
2
 2n  1
8
n
(vii) (2n  7)  (n  3)2 .
12.
If 1  ai  0 for all i, prove that 1  a1 1  a2  ....... 1  an   1  a1  a2 ......an
13.
14.
For any x > -1, show that (1 + x)n ≥ 1 + nx.
Prove by mathematical induction .
(i)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(x)
(xi)
(xii)
(xiii)
(xiv)
2
 n(n  1) 
13  23  33  ....  n3  
 .
2 

1
1
1
2n
1

 ... 

.
(1  2) (1  2  3)
(1  2  3  ....n) (n  1)
1  3  32  ...  3n1 
(3n  1)
.
2
(ii)
n(n  1)(n  2)(n  3)
4
n 1
(2n  1)3  3
1.3  2.32  3.33  ....  n.3n 
4
 n(n  1)(n  2) 
1.2  2.3  3.4  ....  n.(n  1)  

3


2
n(4n  6n  1)
1.3  3.5  5.7  .....  (2n  1)(2n  1) 
3
1.2.3  2.3.4  ....  n(n  1)(n  2) 
1.2  2.22  3.22  ....  n.2n  (n  1) 2n1  2.
(ix)
1 1 1
1
1
   .....  n 1  n .
2 4 8
2
2
1
1
1
1
n


 ..... 

.
2.5 5.8 8.11
(3n  1)(3n  2) (6n  4)
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)
a(r n  1)
.
r 1
 3   5  7   (2n  1) 
2
1   1  1   .....1 
  (n  1)
2
n
 1   4  9  

 1  1   1   1 
1  1   1   ....1    (n  1).
 1  2   3   n 
a  ar  ar 2  ....  ar n 1 
9
(xv)
(xvi)
(xvii)
(xviii)
(xix)
(xx)
n(2n  1)(2n  1)
.
3
1
1
1
1
n


 ....... 

.
1.4 4.7 7.10
(3n  2)(3n  1) (3n  1)
1
1
1
1
n


 ....... 

.
3.5 5.7 7.9
(2n  1)(2n  3) 3(2n  3)
12  32  52  .....  (2n  1)2 
1
1
1
n

 ....... 

1 2 2  3
n   n  1 n  1
2.5 + 5.8 + … upto n terms = n(3n2 + 6n + 1)
7 + 77 + 777 + …………… + 7777……..7( n digits ) =
7
10n1  9n  10 
81
n2
 2n  3  ,
4
n2
2
2
2
2
(xxii) 1 + 2.2 + 3 + 2.4 + …………. upto n terms =
 n  1
2
nx
 n  1
sin 
 x sin
2
 2 
(xxiii) sin x  sin 2 x  sin 3 x  .....  sin nx 
x
sin
2
n
sin 2 x
(xxiv) cos x cos 2 x cos 4 x........ cos( 2 n 1 x)  n
2 sin x
(xxi)
15.
16.
13 – 23 + 33 – 43 + ……………. + (-1)n-1 n3 = -
(n even)
(n odd)
If u1 = u2 = 1, un = un-1 + un-2 (n > 2) prove that
(i)
un+3 = u1 + u2 + ……………. + un+1 + 1 (ii)
u 22n = u1 u2 + u2 u3 + ……… + u2n-1 u2n
(iii)
un2  un 1un 1   1
u2n+1 = 1 + u2 + u4 + ………. + u2n
(v)
un is divisible by 5 if and only if n is divisible by 5.
n 1
(iv)
Prove that if p is a prime number then a p  a is divisible by p for all positive integers
a. (You may use the fact that p C m is divisible by p if p is a prime and m  1, 2,3........ p  1 )
 3n ! is an integer for all
n.
17.
Prove that
18.
Show that n 4  4n  11 is divisible by 16 if n is an odd number of the type 4k  1 .
19.
Show that 22  1 has least n distinct prime divisors.
20.
Prove that if a is odd then a 2  1 is divisible by 2n 2.
21.
Show that u1  3, u2  5 and un1  3un  2un 1. Prove that un  2n  1 for all n.
6n n !
n
n
10
n
 1 5 
if n  1, show that un  
 2  for all n.


22.
If u1  u2  1, un  un 1  un  2
23.
If x1  x2  6, x1 x2  1 and un  x1n  x2n show that un  6un1  un2. Hence show that un is an
integer for all n. Further show that un is not divisible by 5 for any n. Does the result hold true
for a negative integer as well ?
24.
If x  y  u  v and x 2  y 2  u 2  v 2 . Prove that x n  y n  u n  v n for n.
25.
Show by induction :
(i)
1
x 1
x
1
x
1
x
tan  2 tan 2  ..........  n tan n  n cot n  cot x
2
2 2
2
2
2
2
2
1 nCn

C2 n C3
1 1
1
C1 

 .......... 
 1    ...... 
2
3
n
2 3
n
n 1
n
(ii)
(iii)
26.
n
n
1
1
1
1
1
1
 3
 3
 ........ 


 ....... 
3
2n  1 2  2 4  4
2n
 2n   2n n  1 n  2
Two vessels A and B of equal capacity, A is full of water and B is half-full of wine. B is filled
up from A , then A from B the contents being well stirred each time. This double operation is
repeated n times, prove that in A the ratio of water to wine is then
2  1/ 4  :1  1/ 4  .
n
27.
n

Prove by induction that  n  N  1  2
integers. Also show that bn2 

2 n 1
 an  bn 2 where an and bn are odd positive
an2  1
for all positive integers n.
2


28.
Show by the induction 512 divides 32 n 5  160n 2  56n  243 for all n  N.
29.
If a, A  0 be the arbitrary numbers and let


A
A
A 

a1  1/ 2  a   , a2  1/ 2  a1   ........an  1/ 2  an1 

a
a1 
an1 



a  A  a1  A 
Prove that n


an  A  a1  A 
2n1
n7 5n 2n3
n
is an integer for every positive integer n .
 

7
5
3 105
30.
Prove that
31.
Using induction or otherwise, prove that for any non-negative integers m, n, r and k
11
k
 (n  m)
m0
32.
(r  m)! (r  k  1)!  n
k 

 r 1  r  2 
m!
k!


Using mathematical induction, prove that


tan 1 (1 3)  tan 1 1 7   .....tan 1 1 n2  n  1  tan 1 n  n  2 
33.
If x is not an integral multiple of 2 use mathematical induction to prove that
cos x  cos 2 x  .....  cos nx  cos
34.
n 1
nx
x
x sin cos ec
2
2
2
Let 0  A   for i  1,2.....n . Use mathematical induction to prove that
 A  A2  ....  An 
sin A1  sin A2 .....  sin An  n sin  1
 where n  1 is a natural number.
n



you may use the fact that p sin x  (1  p)sin y  sin  px  1  p  y 
Where 0  p  1 and 0  x, y   
 2n  k 


k   2n  4k  1 n  2 k
Let n be any positive integer. Prove that  
.
2

k  0  2n  k   2n  2k  1


 k 
m
35.

for each non negative integer m  n Here
36.
For every positive integer n prove that
prove that  n 

 n  1   
 
p
q
p
Cq
 4n  1 
 
 2
n
m
2n2m
nm
n2m

n  n  1  4n  2 . Hence or otherwise
4n  1  ,
where  x denotes the greatest integer not exceeding x .
37.
Let a,b,c be positive real numbers such that b2  4ac  0 and let 1  c . Prove by induction that
 n 1 
b
a n2
2
 2a(1   2  ....   n

is well defined and  n1 
n
2
for all n  1,2,..... (Here ‘well-
defined’ means that the dominator in the expression for  n 1 is not zero.)
38.
Use mathematical induction to show that  25
12
n 1
 24n  5735 is divisible by  24  for all n .
2
SOLUTIONS
2.
The result is true for n  3 since 56  53  1 is not divisible by 31 . Let the result be true for
n  m where m  3k then 52 m  5m  1 is not divisible by 31

52 m  5m  1  31m  r where 1  r  30
Now we must show that the result is true for n  m  3
Indeed 52 m6  5m3  1

 56.52m  5m.53  1

 56.31m  56.r  5m6  1  5m.53  1
 56 31m  r  5m  1  5m.53  1


 56.31m  5m3.124  56.r
 56.31m  56.r  5m3 53  1
Since 31m and 124 are divisible by 31, 56.r is not divisible by 31 .
We can say P(m  3) is true. This P(m) 
3.
P(m  3)
It is sufficient to show that 4.6n  5n1  9 is divisible by 20
Let 4.6n1  5n1  9  20m then 4.6n1  5n 2  9


 6. 20m  5n1  9  5n 2  9
 20m  5n1  6  5  45

 20m  45  5n1
 4.6.6n  5n 2  9
 20m  5 9  5n

 20m  5n1  6  5  45
(*)
Now 9  5n is clearly divisible by 4 (Prove by induction)

4.
expression at * is divisible by 20 which complete inductions.
Odd multiplies of 24 are 24,72,120,........166,.....
Let 8.7n  4n2  24(2l  1) (Odd multiple of 24 )


Now 8.7n 1  4n 3  7 48l  24  4n  2  4n 3  336l  168  4n3  7.4n 2  336l  168  4n2 (3)
 336l  3.8.22 n1  168
 336l  24 7  2n 1 
 24 14l   7  2n 1 
Since 7  2n1 is an odd number 24 14l   7  2n 1  is an odd multiple of 24 .
5.
We will first show that 32  1 is divisible by 2n 2 for all n . Indeed the result is true for n  1
n
n1
 
and if 32  2n2.m then 32  1  32
n
n
2
1
13


2
 22 n 4 m2  m.2n3
 2n 2.m  1  1

32  1 is divisible by 2n3
n1
We will now show that 32  1 is not divisible by 2n3 . The result is true for n  1 since 8 is not
n
divisible by 16 . If it is true for n then
32  1  m.2n 2 ( from first part) and P(n) for this
n
question will be 32  1  m.2n 2 where m is odd
n
 
n1
Now 32  1  32
n
2


 m2 .2n 4  m.2n3
 m.2n 2  1  1
1
Since m.2n3 is not divisible by 2n 4 ( m is odd)
32  1 is not divisible by 2n 4 .
n1
16.
Let us apply induction on a (Induction can not be applied on p since no one knows which is the
next prime after p !!)
The result is true for a  1 since 1p  1  0 which is divisible by p then consider
(a  1) p  (a  1)  E on using binomial expansion
E  a p  p C1a p 1  p C2 a p  2  ......  p C p 1a  1  a  1


 a p  a  p C1a p 1  p C2 a p 2  .......  p C p 1a

17.
E is divisible by p .
The result is true for n  1 since
Let

 mp  a multiple of p
3!
is an integer
6!!
(3n)!
(3n  3)!
be an integer, say M (n) then M (n  1)  n 1
n
6 (n  1)!
6 n!
 3n  33n  23n  1
 n  1 6
(3n)!
6n n!

 3n  1 3n  2 
2
. M ( n)
Now  3n  1 3n  2  is product of two consecutive integers

 3n  1 3n  2 
2
is an integer

M (n  1) is also an integer.
NOTE:- The result also follows from the fact that number of ways distributing 3n distinct objects
in n identical boxes keeping 3 objects in each cell must be an integer.
18.
The first odd number of the type 4k  1 is 1 and 14  4  11 is divisible by 16
14
Let n4  4n  11  16l where n is of the form 4k  1
The next odd number of the type 4k  1 is n  4
Now  n  4  4  n  4  11
4
 n4 4 C1n3 .4 4 C2 .n2 .42 4 C3 .n.43 4 C4 .44 4n  16  11

 n4  4n  11  16n3  96n2  256n  256  16 = 16l  16 n3  6n2  16n  17

22.

The result is true for n  4 .
1 5 
1 5
1
,1  
 2 
2


1 5 
Let un 1  
 2 


2
n 1
, un  2
1 5 

 2 


n2
then un  un 1  un  2
1 5 

 2 


n 1
1 5 

 2 


n2
(by induction hypothesis)
24.
1 5 

 2 


n 2
1 5 

 2 


n
1 5
 1 5 
 1  

  2 
 2
 

n 2
 3  5   3  5  1  5 

  



 2   2  2 
2
 The result true for n .
From the relations x  y  u  v and x2  y 2  u 2  v2 we easily get xy  uv
Suppose xn1  y n1  u n1  vn1 , xn  y n  u n  vn (Assuming the result for n  1 and n )



Now  x  y  x n  y n   u  v  u n  v n
25.


xn1  xy n  yxn  y n1  u n1  uvn  vu n  vn1

xy x n1  y n1  x n1  y n1  uv u n1  v n1  u n1  v n1

xn1  y n1  u n1  vn1
(i)
If we assume P(n) to be true It is sufficient to show that P(n  1)





xy  uv , x n 1  y n 1  u n 1  v n 1
1
x 1
x
1
x
1
x
1
x
tan  2 tan 2  ....  n tan n  n 1 tan n 1  n 1 cot n 1  cot x
2
2 2
2
2
2
2
2
2
2
15

LHS

1
x
1
x
cot n  cot x  n1 tan n1
n
2
2
2
2


sin
1  cos 
2
 n

2  sin  2cos 
2



  cot x




1
1

x
cot   tan   cot x,   n
n 
2 
2
2
2
1

1
x
cot  cot x  n cot n1  cot x
n
2
2
2
2
(ii)
Denote LHS by S n and RHS by Tn then we have to show Sn 1  Sn  Tn 1 from n
(iii)
Sn1  Sn 
n 1
1
1
1
1
1
 3
 3
 3
. 

3
3
2(n  1)  1 2  2 4  4 6  6
(2n)  2n (2n  2)  2(n  1)

1
1
1
1
 n
 3
 ...... 

 3
 3

3
(2n)  2n 
 2n  1 2  2 4  4 6  6
2n2  3n  1  2n  3n
1

(2n  3)(2n  1)
2  4(n  1)3  (n  1) 

n 1
n
1


3
2n  3 2n  1 (2n  2)  2(n  1)

1
1
+
(2n  3)(2n  1)
2(n  1) 4(n  1)2  1

1
1

2(n  1)  2(n  1)  1) (2(n  1)  1)
(2n  3)(2n  1)

1
1

(2n  3)(2n  1)
2(n  1) (2n  3)(2n  1)

1
1 
1

1



(2n  1) (2n  3)  2n  2  (2n  1)(2n  2)

1
1
1
1   1
1
1
 1
Tn 1  Tn 

 ........ 




 ..........  

2n 2n  1 2n  2   n  1 n  2
2n 
 n  2 n 1

1
thus Sn 1  Sn = Tn 1  Tn for all n
(2n  1)(2n  2)
26.
16
INITIAL POSITION
The initial position of vessels are shown in the figure. V is the initial amount of water while
V
2
V
of
2
water and wine. But after we fill A again from B i.e. completion of first double operation the
V 1 V 3V
V
amount of water in the first vessel  
and amount of wine 

4
2 22
4
is the initial amount of wine. When we fill up B from A then B contains equal amount
 ratio 
3V 4 3


V 4 1
The result is true for n  1







1
2 

3
4  
1
1
1

4

The status of vessels A and B are shown after k double operations. Wk , N k be the amount of
water and wine respectively in A while in B it should be V  Wk ,
V
 Nk
2
Now  k  1 th double operation starts. The amount of water and wine transferred to B are
1
1
1
Wk , Nk . (The amount left over is also the same) then B , now, contains Wk  V  Wk and
2
2
2
1
V
1
V 1
Nk   Nk which are equal to V  Wk and  N k respectively.
2
2
2
2 2
When again A is filled from B , the amount of water in A after the completion of  k  1 th
double operation
1
1
1 
V 1
 Wk   V  Wk    Vk  Wk 1 and that of wine
2
2
2 
2 4
V 1
  Nk
4 2
 N k 1
V 1
 Vk
Wk 1
 2 4
Thus
N k 1 V  1 N
k
4 2
17


1
V 1
Nk   Nk
2
4 4
2V  Wk 2 Wk  N k   Wk

V  Nk
Wk  N k  N k
Wk 1
N k 1

Wk
2
3Wk  2 N k
Nk


Wk
Wk  2 N k
2
Nk
3.
n
n 1
n
1
1
1
2 
8 
2 
W
4 
 4  we easily get Wk 1 
4
But by hypothesis k 
n
n
n
Nk
N k 1
1
1
1
4 
1  
1  
4
4
4
27.
1  2 

2
75 2
Let 1  2

2 n 1

Since 7 and 5 are odd the result is true for n  1

 an  bn 2 where an , bn are odd then 1  2


2 n 3

 1 2
 1  2 
2 n 1

 an  bn 2 3  2 2   3an  4bn    3bn  2an  2
 an1  bn1 2
Since an 1 and bn 1 are odd (Why?)
the result is true for n  1 also.
an 1  3an  4bn ,
To prove the other part use the results
28.
bn 1  3bn  2an
f (n  1)  32 f (n)
 32n7  160(n  1)2  56(n  1)  243 9  32 n5  160n2  56n  243
 8 160n2  768n  8  256

 256 5n2  3n  8

which is a multiple of 512
(since 5n2  3n  5n2  5n  2n  8
 5n(n  1)  2(n  4)
29.
 5.2k  2.(n  4) which is divisible by 2 )
The result is obviously true for n  1
a  A
Let
 1

an  A  a1  A 
an  A
an  A  an  A
an  A  an  A
2n1
 t 2 for some n then
n1

t2 1
n1
t2 1

18
an  A.
1 t2
n 1
1 t2
n 1
2
To complete the induction we must show that
1
A
 an    A
2
an 
Now LHS  
1
A
 an    A
2
an 
 an  A 


 an  A 
2
 an

1 

A


 an  1 


 A


an1  A
an1  A
n
an 2  2 A an  A
an 2  2 A an  A
 1 t2

1 

2 n1

  1  t 2n1
 1 t

1

2 n1
 1 t

n 1
2
 t2
2
 t2  t2

 2t 2n1

n1
n1
2

2n
  t

MATHEMATICAL INDUCTION ( NCERT SOLUTIONS)
16.
For n  1,
LHS 
1
1

1.4 4
Let P  k  be true then
RHS 
1
1

3 1  1 4
1
1
1
1
k


 .......

1.4 4.7 7.10
 3k  2 3k  1 3k  1
(i)
To prove P  k  1 we must show that
1
1
1
1
1
k 1


 ...... 


1.4 4.7 7.10
 3k  23k  1 3k  13k  4  3  k  1  1
(Note that the extra term is being written by pattern i.e. adding 3 or by replacing k by k  1 )
LHS 
1
1
1
1

 ..... 

1.4 4.7
 3k  2 3k  1 3k  13k  4 

k
1
from P  k 

3k  1  3k  1 3k  4 

1 
1 
k

3k  1 
3k  4 
1  3k 2  4k  1

3k  1  3k  4 
19

 3k  1 k  1
 3k  1 3k  4 

18.
k 1
3k  4
 RHS
The result is true for n  1 since 1 
1
9
2
 2 1  1 i.e. 1 
8
8
Let P  k  be true then 1  2  3  .....  k 
1
2
 2k  1 (i)
8
1
2
 2k  3 
8
(ii)
1
2
 2k  1   k  1 by
8
(iii)
We must show that 1  2  3  ......  k   k  1 
we have 1  2  3  .....  k   k  1 
From (iii) the statement (ii) follows if
 2k  1
2
 8  k  1   2k  3
2
1
1
2
2
 2k  1   k  1   2k  3 or
8
8
or 4k 2  12k  9  4k 2  12k  9 which is, indeed true
for all k .Thus P  k   P  k  1
19.
Let k  k  1 k  5  3m
We must show that  k  1 k  2 k  6  is also a multiple of 3 .
Now  k  1 k  2 k  6    k  1 k  2 k  6  k  1 k  2
(Avoid touching 6  k  1 k  2 ) since it is a divisible by 3 )
 k  k  1 k  5  3  3m  k  k  1 k  5  3k  k  1  3m
 3m  3k  k  1  3m which is divisible by 3 Thus P  k 
21.
 P  k  1
A polynomial f  x  is divisible by another polynomial g  x  if there exist a polynomial
h  x  such that f  x   g  x  h  x 
Let x2k  y 2k   x  y  h  x  where h  x  is a polynomial.
20
Now x 2 k  2  y 2 k  2  x 2 k .x 2  y 2 k 2  x 2  x  y  h  x   y 2 k   y 2 k  2
  x  y  x2 h  x   x 2 y 2k  y 2k 2  x 2  x  y  h  x   y 2 k  x 2  y 2 
 x2  x  y  h  x   y 2k  x  y  x  y  which is also divisible by x  y
Thus P  k 
22.

P  k  1
Let 32 k  2  8k  9  8m
Now 32k 4  8  k  1  9  32.32k 2  8  k  1  9
 32 8m  8k  9  8  k  1  9  72m  72k  81  8k  8  9
 72m  64k  64

23.
P  k  1 is also true since the last expression is divisible by 64
 IH 
Let 41k  14k  27 m
Now 41k 1  14k 1
 41.41k  14.14k
 41 27m  14k   14.14k
 41.27m  41.14k  14.14k
 27m  14k.  41 14 
 27 m  14k .27

24.
P  k  1 is also true since the last expression is divisible by 27
 2n  7    n  3 
2
(i)
The result is true for n  1 since 9  16
Let the result be true for n  k then 2k  7   k  3
2
(ii)
To prove the result for n  k 1 we must show that 2  k  1  7   k  4 
21
2
(iii)
Now 2  k  1  7   2k  7   2   k  3  2 it is now sufficient to show
2
 k  3  2   k  4  or
P k  
P  k  1
2
2
6k 13  8k 16
22
or
2k  3  0 which is true thus