Chapter 2 Mathematical Induction

Advanced Mathematics Course
Chapter 2 Mathematical Induction
Advanced Mathematics Course
Chapter 2 Mathematical Induction
Principle of Mathematical Induction
For any n  , let P(n) be a statement on n. If it can be proved that
(i) P(1) is true, and
(ii) P(k) is true  P(k + 1) is true, where k 
then the statement P(n) is true  n  .
,
Theorem 2.1 (Variations of Mathematical Induction)
For any n  , let P(n) be a statement on n.
(a) Let r  . If it can be proved that
(i) P(r) is true, and
(ii) P(k) is true  P(k + 1) is true, where k 
with k ≥ r,
then the statement P(n) is true  n 
with n ≥ r.
(b)
(Two-step induction)
Let r  . If it can be proved that
(i) P(r) and P(r + 1) are true, and
(ii) P(k) and P(k + 1) are true  P(k + 2) is true, where k 
then the statement P(n) is true  n 
with n ≥ r.
(c)
with k ≥ r,
(Jumped induction)
Let r, s  . If it can be proved that
(i) P(r) is true, and
(ii) P(k) is true  P(k + s) is true, where k 
with k ≥ r,
then the statement P(r), P(r + s), P(r + 2s), P(r + 3s), … are all true.
(d)
(Backward induction)
Let r  . If it can be proved that
(i) P(r) is true, and
(ii) P(k) is true  P(k – 1) is true, where k 
with k ≤ r,
then the statement P(n) is true  n 
with n ≤ r.
Example 2.1
Prove that 32n – 32n2 + 24n – 1 is divisible by 512 n 
.
Example 2.2
Prove that n! > 2n n 
with n ≥ 4.
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Advanced Mathematics Course
Chapter 2 Mathematical Induction
Example 2.3
Prove that 1·n + 2(n – 1) + 3(n – 2) + … + (n – 1) ·2 + n·1 =
1
n(n  1)(n  2) n 
6
.
Example 2.4
Let {Fn} be a sequence with F1 = 0, F2 = 1 and Fn+2 = Fn+1 + Fn n 
 
n
. Prove that
n
n 
5
where  >  and ,  are roots of the quadratic equation x2 – x – 1 = 0.
Fn 
Example 2.5
Let a 
+
. Prove that a n  a n  2  a n  4  ... 
1
a
n4

1
a
n2

1
 n  1 n 
an
.
Example 2.6
Let {pn} and {qn} be two sequences of positive numbers such that
(i) p1 > q1,
(ii)
(iii)
pn 1 
1
( pn  qn ) ,
2
pn qn  c n 
, where c is a positive constant.
Prove that pn > pn+1 > c > qn+1 > qn n 
.
Example 2.7*
Let {an} be a sequence of positive numbers.
Let An 
(a)
(b)
(c)
a1  a2  ...  an
and Gn  n a1a2 ...an n 
n
.
Prove that An ≥ Gn for n = 2, 4, 8, 16, 32, …, 2k, … where k  .
Prove that if Ar ≥ Gr for some r 
and r ≥ 2, then Ar–1 ≥ G r–1.
Prove that An ≥ Gn n  .
Remark:
The above inequality is the famous “AM-GM inequality”, where An is called the arithmetic
mean (AM) and Gn is called the geometric mean (GM).
Example 2.8
Let a1, a2, a3, …, an be n positive numbers.
(a) Prove that if a1·a2· … ·an = 1, then a1 + a2 + … + an ≥ n.
n
(b)
Using (a), prove that
a
r 1
n
r

n
n
a
r 1
r
.
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Advanced Mathematics Course
Chapter 2 Mathematical Induction
Second Principle of Mathematical Induction
Theorem 2.2 (Second Principle of Mathematical Induction)
For any n  , let P(n) be a statement on n. If it can be proved that
(i) P(r) is true, and
(ii) P(r), P(r + 1), P(r + 2), …, P(k) are all true  P(k + 1) is true, where k 
then the statement P(n) is true  n 
with n ≥ r.
with k ≥ r,
Remarks:
The second principle differs from the original principle in the induction step. In the induction

assumption, it is assumed that all the previous cases are true.
The principle works by the following argument:

By (i), P(1) is true.
∵ P(1) to P(1) are (should be “is” here) true, by (ii), P(2) is true.
∵ P(1) to P(2) are true, by (ii), P(3) is true.
∵ P(1) to P(3) are true, by (ii), P(4) is true.

∵ P(1) to P(4) are true, by (ii), P(5) is true.
…
The second principle is sometimes called “strong induction” because of the strong assumption.
Example 2.9
Let {an} be a sequence of positive numbers such that
 1  an 
a1  a2  ...  an  

 2 
Prove that an = 2n – 1  n 
Exercises
1.
Prove that
2.
n
3.
n
.
.
n33 n
with n ≥ 4.
Let {an} be a sequence with a1 = 1, a2 
Prove that an 
2
1
n
2n 1
1
and 2(n + 2)an+2 – 3nan+1 + (n – 1)an = 0  n 
2
.
.
Let   1  3 and   1  3 . Let xn 
xn and yn are integers and yn is even  n 
1
2 3
( n   n ) and yn =  n +  n  n 
. Prove that
.
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Advanced Mathematics Course
Chapter 2 Mathematical Induction
Solution to Exercises
1.
When n = 4, 4 4  2  6 8 and
∴ the result is true for n = 4.
3
3  6 9  6 8.
Assume the result is true for n = k for some k 
i. e. k k  3 3  k3 < 3k
Consider
(k + 1)3 – 3k+1
=
k3 + 3k2 + 3k + 1 – 3·3k
<
k3 + 3k2 + 3k + 1 – 3k3
=
–2k3 + 3k2 + 3k + 1
≤
–2(4k2) + 3k2 + 3k + 1
=
–5k2 + 3k + 1
≤
=
≤
∴
∴
2.
with k ≥ 4.
(by induction assumption)
(∵ k ≥ 4)
–5(4k) + 3k + 1
–17k + 1
0
(k + 1)3 < 3k+1
(∵ k ≥ 4)
(taking 3(k + 1)-th root on both sides)
∴
k 1  3 3
the result is also true for n = k + 1.
∴
by the principle of mathematical induction, the result is true  n 
(∵ k ≥ 4)
k 1
When n = 1, LHS = 1, RHS =
∴
1
= 1.
211
the result is true for n = 1.
When n = 2, LHS =
∴
1
1
1
, RHS = 21  .
2
2
2
the result is true for n = 2.
Assume the result is true for n = k and n = k + 1 for some k 
i. e.
.
ak 
.
1
1
1
and ak 1  k 11  k
k 1
2
2
2
Then,
2(k + 2)ak+2 – 3kak+1 + (k – 1)ak = 0
2(k  2)ak  2  3k
1
1
 (k  1) k 1 = 0
k
2
2
2(k + 2)ak+2 =
=
ak+2 =
(by induction assumption)
1
(3k  2(k  1))
2k
1
( k  2)
2k
1
2 k 1
∴
the result is also true for n = k + 2.
∴
by the principle of mathematical induction, the result is true  n 
.
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Advanced Mathematics Course
3.
1  3 1  3
 1 , yn      1  3  1  3  2 .
2 3
2 3
x1 and y1 are integers and y1 is even.
When n = 1, x1 
∴
∴
Chapter 2 Mathematical Induction
1
(   ) 
the result is also true for n = 1.
Assume that the result is true for n = k for some k 
i. e. xk and yk are integers and yk is even.
Then,
xk 1 
1
( k 1   k 1 )
.
yk 1   k 1   k 1
2 3
1

(   k     k )
2 3
1

((1  3) k  (1  3)  k )
2 3
1

( k   k  3( k   k ))
2 3
1
 xk  yk
2
Since xk and yk are integers and yk is even, and xk 1  xk 
   k     k
 (1  3) k  (1  3)  k
  k   k  3( k   k )
 yk  3(2 3
k k
2 3
)
 yk  6 xk
1
yk , xk+1 is an integer.
2
Since xk and yk are integers and yk is even, and yk+1 = yk + 6xk, yk+1 is an integer and is even.
∴ the result is also true for n = k + 1.
∴
by the principle of mathematical induction, the result is true  n 
.
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