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
n4
1
a
n2
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
.
.
n33 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.
211
the result is true for n = 1.
When n = 2, LHS =
∴
1
1
1
, RHS = 21 .
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 11 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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