Mathematical Induction
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
3
Algebra
Chapter 3
Mathematical Induction
3.1
First Principle of Mathematical Induction
2
3.2
Second Principle of Mathematical Induction
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Mathematical Induction
Advanced Level Pure Mathematics
3.1 First Principle of Mathematical Induction
Definition
Let P (n ) be a proposition on an integer variable n. Then P (n ) is true for all integers n  s
if and only if the following two conditions are both satisfied :
Example 1
(i)
P(s) is true ,
ii)
If P (k ) , where k  s , is assumed to be true, then P (k  1) is true .
Prove that for all positive integers n ,
(13  2 3  33    n 3 )  3(15  2 5  35    n 5 )  4(1  2  3  n) 3
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Example 2
Prove, by induction, that
1
1
1
1 1 1
1
1

 ... 
 1     ... 

n 1 n  2
2n
2 3 4
2n  1 2n
for all positive integers n.
(1)
When n = 1,
L.H.S. =
1
2
R.H.S. = 1 
1 1

2 2
Hence, the proposition is true for n = 1.
(2)
Assume the proposition is true for n = k , k  1.
i.e.
1
1
1
1 1 1
1
1

 ... 
 1     ... 

k 1 k  2
2k
2 3 4
2k  1 2k
When n = k + 1,
L.H.S. =
1
1

 ... 
k 1 k  2
1

1

1
=
Hence, the proposition is also true for n = k + 1.
By the Principle of Mathematical Induction, the proposition is true for all
positive integers n.
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Mathematical Induction
Advanced Level Pure Mathematics
Example 3
A sequence a1 , a2 ,  , an is defined as follows :
a1 = 1 , a2 = 2 and an+2 = an+1 + an for n  1.
Prove, by induction, that an+12  an an+2 = (1)n1 for all positive integers n.
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Mathematical Induction
Advanced Level Pure Mathematics
Example 4
Given a sequence u1 , u 2 ,, u n , such that u1  1 and u n  u n1  3 , ( n  2) .
Show that u n  3n  2 , for all positive integers n .
Example 5
Prove that for any positive integer n , there exist unique positive integers an
and bn such that (1+ 3 )n = an + bn 3 .
(1)
Want to prove the existence of integers an and bn .
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Mathematical Induction
Advanced Level Pure Mathematics
(2)
Want to prove the uniqueness of integers an and bn .
Example 6
Prove, by induction, that n(n2 + 5) is divisible by 6 for all positive integers n.
(1)
When n = 1,
n(n2 + 5) = 16 = 6
Hence, the proposition is true for n = 1.
(2)
Assume the proposition is true for n = k , k  1
i.e.
k(k2 + 5) = 6m where m is an integer.
When n = k + 1,
n(n2 + 5) = (k+1)[(k+1)2 + 5]
[ Hint : For any integers n ,
n(n  1)
is also an integer. ]
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Mathematical Induction
Advanced Level Pure Mathematics
Modification of The First Principle of Mathematical Induction
Definition
If a proposition P(n) is true for a positive odd/even integer s and that P(k) is true implies
P(k+2) is also true, then P(n) is true for all positive odd/even integers n  s.
Example 7
Prove, by mathematical induction, that 5 n  2 n is divisible by 21 for all positive even
integers n.
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Mathematical Induction
Advanced Level Pure Mathematics
Example 8
Prove, by mathematical induction, that 5 n  3 n  2 n is divisible by 30 for all positive odd
integers n greater than 1.
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Mathematical Induction
Advanced Level Pure Mathematics
3.2 Second Principle of Mathematical Induction
Definition
Let P(n) be a proposition involving positive integers n. If
(i)
P(n) is true for n = 1 and n = 2 , and
(ii)
if P(n) is true for some positive integer k and k+1, then P(n) is also true for
n = k+2, then P(n) is true for all positive integers n.
Example 9
A sequence of real numbers a1 , a2 ,  , an ,  is defined by a0 = 0 , a1 = 1 and
an+2 = an+1 + an for all n = 0 , 1 ,  .
Show that for all non-negative integers n ,
an 
1
5
( n   n )
where  ,  are roots of the equation x2 + x  1 = 0 with  >  .
[ Hint :  is a root of the equation x2 + x  1 = 0   2 +   1= 0. ]
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Mathematical Induction
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Example 10 A sequence of real numbers a0 , a1 ,, an , is defined by
a0  1, a1  7 and an 2  4an1  3an  0 for n  0,1,2, 
Prove, by induction, that a n  3 n 1  2 for all non-negative integers n .
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Mathematical Induction
Advanced Level Pure Mathematics
Example 11 Consider the sequence {Un } in which
U1 = 0 , Un +1 = 2n  Un for n = 1, 2,  .
Using mathematical induction or otherwise, show that
2Un = 2n  1 + (1)n for n = 1, 2,  .
Un
n  n
Hence find lim
[HKAL92]
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Mathematical Induction
Advanced Level Pure Mathematics
Example 12 Let U1 = 1 , U2 = 3 and Un = Un2 + Un1 for n  3 .
Using mathematical induction, or otherwise, prove that
Un =  n +  n for n  1 ,
where  and  are the roots of x2  x  1 = 0 .
Example 13
[HKAL93]
Let {an} be a sequence of positive numbers such that
 1  an 
a1 + a2 +  + an = 
 for n = 1, 2, 3,  .
 2 
2
Prove by induction that an = 2n  1 for n = 1, 2, 3,  .
[HKAL94]
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Mathematical Induction
Advanced Level Pure Mathematics
Example 14 Let {an } be a sequence of real numbers, where it is defined by
a0 = 1 , a1 = 6 , a2 = 45 and
an  an+1 
1
1
an+2 
an+3 = 0 for n = 0, 1, 2,  .
3
27
Using mathematical induction, or otherwise, show that
an = 3n(n2 + 1) for n = 0, 1, 2, 
[HKAL97]
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Mathematical Induction
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Example 15 Let  ,  be the roots of x 2  14 x  36  0 .
Show that  n   n is divisible by 2 n for n  1,2,3,
Example 16
A sequence {an} is an increasing sequence of positive numbers. Suppose sequences
{bn} , {cn} are defined such that
b1 = c1 =
1
1
a1 , bn = (an1 + cn1) and cn = an 1bn1 for n  2 .
2
2
Show by induction that
(a)
(a)
bn < bn+1 and cn < cn+1 .
(b)
bn < an and cn < an for n  1 .
Want to prove bn < bn+1 and cn < cn+1.
When n = 1
b2 =
1
(a1 + c1)
2
=
1
1
(a1 + a1)
2
2
=
3
a1
4
>
1
a1 = b1
2
c2 = a1b1
=
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Mathematical Induction
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Hence, the proposition is true for n = 1.
Assume bk < bk+1 and ck < ck+1 for k  1.
When n = k+1,
bk+2  bk+1
=
1
1
(ak+1 + ck+1)  (ak + ck)
2
2
=
1
1
(ak+1  ak) + (ck+1  ck)
2
2
ck+2  ck+1
=
>0
 bk+1 < bk+2
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Mathematical Induction
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Example 17 A sequence of real numbers {an} is defined as follows :
a0 = a1 = 1 , a2 = 3 and an+3 = 3an+2  an+1  2an for n = 0, 1, 2,  .
(a)
Let bk = ak+2  ak+1  ak for k = 0, 1, 2,  .
Prove that bn = 2bn1 for all n  N.
(b)
Hence, or otherwise, deduce that an  2n for all n = 0, 1, 2,  .
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