IB Mathematics HL
proof by mathematical induction - 1
n
1.
Prove that
1
 r  r  1  3 n  n  1 n  2
r 1
2.
Prove that n  n  1 n  2  is divisible by 6 for all n such that n 
3.
(a) Find the first three derivatives of xe x .
(b) Suggest a formula for the nth derivative of xe x , that is

.
d
xe x  , n 

dx

.
(c) Prove that your formula is true by mathematical induction.
1
IB Mathematics HL
proof by mathematical induction - 1
Solutions
1.
(i) Show statement true for n  1 :
1
1
 r  r  1  3 11  11  2
r 1
1
1
1
11  11  2    2 3  2  statement true for n  1
3
3
r 1
(ii) Assume statement is true for a specific value of n   ; i.e. statement true for some n  k :
k
1
that is, assume  r  r  1  k  k  1 k  2 
3
r 1
(iii) Show that it must follow from this assumption that the statement is true for the next value of n;
that is, show statement must be true for n  k  1 :
k 1
1
r  r  1   k  1 k  2  k  3

3
r 1
sum of k  1 terms = sum of k terms + value of term when r  k  1
k
1
r  r  1   k  1 k  2    k  1 k  2  k  3

3
r 1
1
1
k  k  1 k  2    k  1 k  2    k  1 k  2  k  3 [applying assumption from (ii)]
3
3
1
1
 k  1 k  2  k  1   k  1 k  2  k  3
3
 3
1
1
Q.E.D.
 k  1 k  2  k  3   k  1 k  2  k  3
3
3
The statement has been shown true for n  1 and given it’s true for some n  k , n   it has
been shown that it follows that it must also be true for n  k  1 ; therefore, by mathematical
induction the statement is true for all n   .
 r  r  1  11  1  2
2.
and
(i) Show statement true for n  1 : n  n  1 n  2   11  11  2   6 and 6 is divisible by 6, thus
statement is true for n  1 .
(ii) Assume statement is true for a specific value of n   ; i.e. statement true for some n  k :
that is, assume k  k  1 k  2  is divisible 6; which means that k  k  1 k  2  must be equal
to a multiple of 6; i.e. k  k  1 k  2   6 p where p  
(iii) Show that it must follow from this assumption that the statement is true for the next value of n;
that is, show statement must be true for n  k  1 :
need to show that  k  1 k  2  k  3 is a multiple of 6
k 3  6k 2  11k  6
k
3
 3k 2  2k   3k 2  9k  6
6 p  3  k  1 k  2 
[note: k  k  1 k  2   k 3  3k 2  2k ]
[applying assumption from (ii)]
 k  1 k  2 is product of two consecutive integers – thus, one factor is even (a multiple of 2);
therefore the product  k  1 k  2  must be a multiple of 2; i.e.  k  1 k  2   2q, q  
6 p  3  k  1 k  2   6 p  3  2q  6 p  6q  6  p  q 
Q.E.D.
The statement has been shown true for n  1 and given true for some n  k , n 

it’s been shown it follows
that it must also be true for n  k  1 ; therefore, by mathematical induction the statement is true for all n 
© InThinking – IB Maths HL & SL
2

.
IB Mathematics HL
proof by mathematical induction - 1
3. (a) Using the product rule, the first three derivatives of xe x are:
d2
d3
d
x
;

x

2
e
  3  x  e x
 1  x  e x ;


2
3
dx
dx
dx
(b) These results suggest the following formula for the nth derivative in terms of n:
dn
x
  n  x  n  e x
n  xe   1
dx
d
1
xe x    1  x  1 e x  1  x  e x which agrees with

dx
result from part (a), there statement is true for n  1 .
(c) (i) Show statement true for n  1 :
(ii) Assume statement is true for a specific value of n 
dk
k
that is, assume
xe x    1  x  k  e x
k 
dx

; i.e. statement true for some n  k :
(iii) Show that it must follow that the statement is true for the next value of n; that is, show
statement must be true for n  k  1 :
d k 1
k 1
xe x    1  x   k  1  e x
k 1 
dx
the  k  1 derivative = derivative of the kth derivative

d  dk
x
  k 1  x  k  1 e x

k  xe    1
dx  dx

d 
k
1  x  k  e x   RHS


dx 
 1
k
 1
k


d
x
x 
 dx  xe  ke    RHS
d
d
x
x 
 dx  xe   dx  ke   RHS
 1 1  x  e x  ke x   RHS
k
 1  1 x  k  1 e x   RHS
k
 1  1  x  k  1 e x   RHS
k
 1  x  k 1 e x   1  x  k  1 e x
k 1
k 1
Q.E.D.
The statement has been shown true for n  1 and given it’s true for some n  k , n   it has
been shown that it follows that it must also be true for n  k  1 ; therefore, by mathematical
induction the statement is true for all n   .
© InThinking – IB Maths HL & SL
3