Section Check In – Core Pure: Proof
Questions
1.*
Prove by induction that 5.3n -1 is divisible by 2 for n
2.
Prove by mathematical induction that
.
  r 2 + 2r  = 6 (n +1)(2n + 7) is true for all n 
n
n

.
r 1
 1 1
n  1 n  for n .

 . Prove by induction that M = 
0 1 
0 1 
3.
Let the matrix M  
4.
A sequence is given by xn1  xn  5 , x1  1 . Prove by induction that xn = 5n - 4 .
5.*
Prove that the nth derivative of xe x is (1) n ( x  n)e x for n 
6.
The Fibonacci numbers are defined as follows. F1  F2  1 , Fn  Fn1  Fn2 for n  2 .
n
Prove that
F
r 1
7.*
2
r
.
 Fn Fn 1 .
(i) Show that 102n  2 is not always divisible by 5 and 7 for n 
(ii) Prove that 102n  2 is divisible by 3 for all n 

Prove by mathematical induction that

.
.
  r  r 3    4 (n2 1)(n  2) is true for all n 
n
8.

n

.
r 1
9.*
Prove that 2n  2n1  2n1  1 for positive integers n.
10.*
Prove that n !  2 n for positive integers n  4 .
Extension
n
n
n
n
Let n  6 be an integer. Show that    n !    .
3
 
2
n
 n 1
f (n)  e .]
[You may use the result that f (n)  
 is an increasing function and nlim

 n 
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Worked solutions
1.
Consider n  1
5.31  1  5(3)  1  15  1  14  2(7) divisible by 2 .
Therefore true for n  1 .
*Assume true for n  k
Let f( k )  5.3k  1
Then f ( k ) is assumed divisible by 2 and so f (k )  2m for some m
.
*Consider n  k  1
f (k  1)  f (k )  5.3k 1  1  5.3k  1  5.3.3k  5.3k  10.3k  2.5.3k
so f (k  1)  f ( k ) is divisible by 2 .
So f (k  1)  f (k )  2l for some l 
hence f (k  1)  f (k )  2l  2(m  l ) which is divisible by 2 .
In conclusion
The statement is true for n  1 . If assumed true for n  k the result has then been shown to
be true for n  k  1 . Therefore, by mathematical induction the result is true for all n .
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2.
Consider n  1
LHS is 12  2(1)  1  2  3
1
1
18
(1  1)(2(1)  7)  (2)(9)   3
6
6
6
LHS  RHS
Hence true for n  1 .
RHS is
Assume true for n  k , so
  r 2  2r   6 (k  1)(2k  7) assumed true.
k
k
r 1
Consider n  k  1
k 1

r 1


k


r 2  2r   r 2  2r  (k  1)2  2(k  1)
r 1
k
(k  1)(2k  7)  ((k  1) 2  2(k  1))
6
Taking out a factor of
k 1
6
(k  1)
(k  1)
(k  1)
2k 2  13k  18 
2k  9)(k  2
 k (2k  7)  6(k  1)  12  
6
6
6
(k  1)

  k  1  1)(2(k  1)  7  .
6





The statement is true for n  1 . If assumed true for n  k the result has then been shown to
be true for n  k  1 . Therefore by mathematical induction the result is true for all n 
3.

.
If n  1
 1 1
M1  
 which is true
0 1 
 1 k 

0 1 
Assume true for n  k , so that M k  
Consider n  k  1
 1 k  1 1  1  0 1  k   1 (k  1) 
M k 1  



 Therefore true when n  k 1 .
1 
 0 1  0 1   0  0 0  1   0
The statement is true for n  1 . If assumed true for n  k the result has then been shown to
be true for n  k  1 . Therefore by mathematical induction the result is true for all n .
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4.
Basis case
When n  1 we have x1  5(1)  4  5  4  1 and x1  1 is given, therefore true when n  1 .
Inductive step
Assume true when n  k therefore xk  5k  4 assumed true.
Consider n  k  1
xk 1  xk  5  5k  4  5  5(k  1)  4 so true for n  k 1 .
If assumed true for n  k the result has then been shown to be true for n  k  1 .
Therefore by mathematical induction the result is true for all n 
5.

.
Basis case
When n  1
d
xe x  e x  xe x  (1  x)e x  ( x  1)e x  (1)1 ( x  1)e x so true for n  1
dx


Inductive step
Assume true for n  k , so that
Consider n  k  1


dk
xe x  (1)k ( x  k )e x
k
d x
d k 1
d  dk
d
x
x 
k
x
x
e

 k xe   (1) ( x  k )e 
k 1
d
x
d
x
d x
d x





 (1)k (1)e x  (1)k ( x  k )(e x )  (1)k e x  ( x  k )e x 
 (1)k (1) ( x  k  1)e x   (1)k 1 ( x  (k  1))e x  so true for n  k  1
If assumed true for n  k the result has then been shown to be true for n  k  1 .
Therefore by mathematical induction the result is true for all n 
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6.
Consider n  1
1
F
r 1
2
r
 1  F1 F2 hence true for n  1
Assume true for n  k so
k
F
r 1
2
r
 Fk Fk 1 assumed true.
Consider n  k  1
k 1
k
 Fr2   Fr2 Fk21  Fk Fk 1  Fk21
r 1
r 1
k 1
So
F
r 1
2
r
 Fk Fk 1  Fk21  Fk 1  Fk  Fk 1   Fk 1Fk  2
Therefore true for n  k  1 .
The statement is true for n  1 .
If assumed true for n  k the result has then been shown to be true for n  k  1 . Therefore
by mathematical induction the result is true for all n 
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7.
The values of 102n  2 for the first few natural numbers are 102, 10 002, 1 000 002, …
These are not divisible by 5 because of the final digit 2 .
102 is not divisible by 7 3 .
Basis case
Consider n  1
1021 + 2 =102 + 2 =100 + 2 =102 = 34  3divisible by 3 .
Therefore true for n  1 .
Inductive step
Assume true for n  k
Therefore 102n  2 assumed divisible by 3 .
Consider n  k  1
Now let f (k )  102 k  2
and consider the difference f (k  1)  f ( k )  102( k 1)  2  102 k  2
 102k  2  102k  102k .102  102 k  100.102 k  102 k  99.102 k  3  33 102 k
So f (k  1)  f ( k ) is divisible by 3 .
Now f (k  1)  f ( k )  3  33 102 k , and both f ( k ) and 3 33 102k are divisible by 3 .
Hence f ( k  1) is divisible by 3 .
If assumed true for n  k the result has then been shown to be true for n  k  1 . Therefore
by mathematical induction the result is true for all n 
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8.
Consider n  1
LHS  1  (1)3  1  1  0
1
1
RHS   (12  1)(1  2)   (0)(3)  0
4
4
LHS  RHS
Hence true for n  1 .
Assume true for n  k , so
  r  r 3    4 (k 2  1)(k  2) assumed true.
k
k
r 1
Consider n  k  1
k 1
r
r 1
 r3
  r
k
r 1
 r3

 (k  1)  (k  1)3
k
  (k 2  1)(k  2)  (k  1)  ( k  1)3
4
k
  (k  1)(k  1)(k  2)  (k  1)  (k  1)3
4

(k  1)
k (k  1)(k  2)  4  4(k  1) 2
4

(k  1)
k (k  1)(k  2)  4k 2  8k
4

(k  1)
 k{(k  1)(k  2)  4(k  2)}
4

(k  1)
 k (k  2)((k  1)  4) 
4

(k  1)
(k  1) 2  1)((k  1)  2)
4






Therefore true for n  k  1 .
If assumed true for n  k the result has then been shown to be true for n  k  1 . Therefore
by mathematical induction the result is true for all n 
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9.
When n  1 2n  2 and 2n 1  2n 1  1  4  1  1  2 , so we have equality.
Hence true for n  1 .
Assume true for n  k , so that 2k  2k 1  2k 1  1
Then for n  k  1 ,


2k 1  2  2k  2  2k 1  2k 1  1  2( k 1)1  2( k 1)1  2  2( k 1)1  2( k 1)1  1
so true for n  k  1 .
If assumed true for n  k the result has then been shown to be true for n  k  1 . Therefore
by mathematical induction the result is true for all n 
10.

.
For this proof by induction, the first value of n is 4.
When n  4 , 4! = 24 and 2 4  16 so n !  2 n .
Assume true when n  k , so that k !  2 k assumed true.
Consider n  k  1
 k  1!  (k  1)k !  (k  1)2k  2k 1 since
k 1  2
so true for n  k  1 .
If assumed true for n  k the result has then been shown to be true for n  k  1 . Therefore
by mathematical induction the result is true for all n 

such that n  4 .
Extension
(NB some explanatory steps have been left out – make sure you can justify each step)
Basis case
6
6
6
6
3
2
For the inductive step we show each half of the inequality separately for clarity.
When n  6 we have    6!    which evaluates as 64  720  729 so it is true.
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n
n
1. First    n !
3
k
k
Assume true for n  k so that    k !
3
Then for n  k  1 we have
 k 1 


 3 
k 1
k
k
k
k
 k   k 1 k 1
 k 1 k 1
 k 1  1
   
 k ! 
 (k  1)! 
 
 
 
3
3
3  k 
 k 
 k  3
n
 n 1 
f (n)  e , we can see
 is an increasing function and nlim

 n 
Since we can assume that f ( n)  
k
 k 1  1
 k 1
   1 so 

 k  3
 3 
that 
n
2
k 1
 (k  1)! as required.
n
2. Now n !   
k
In a similar way, assume true for n  k , ie that k !   
2
Then for n  k  1 we have
 k 1 


 2 
k 1
k
k
k
k
k
 k   k 1 k 1
 k 1 k 1
 k 1 1
   
 k ! 
 (k  1)! 
 
 
 
2
2
2  k 
 k 
 k  2
n
 n 1 
f (n)  e , we can see
 is an increasing function and nlim

 n 
Since we can assume that f ( n)  
k
 k 1 1
   1 for all values of k which are sufficiently large.
 k  2
that 
k
6
 k 1  1  7  1
k  6 ; for k  6 , 
       1.26....  1
 k  2 6 2
 k 1

 2 
so 
k
k 1
 (k  1)! as required.
k
k
k
  k !    assumed true for n  k the result has then been shown to be true for n  k 1 .
3
2
Therefore by mathematical induction the result is true for all integers n  6 .
If 
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