Chapter 4
Sequences and Mathematical
Induction
4.2
Mathematical Induction
Mathematical Induction
• A recent method for proving mathematical
arguments.
• De Morgan is credited with its discovery and and
name.
• The validity of proof by mathematical induction is
taken as a axiom.
– an axiom or postulate is a proposition that is not
proved or demonstrated but considered to be either
self-evident, or subject to necessary decision.
(Wikipedia)
Principle of Mathematical Induction
• Let P(n) be a property that is defined for
integers n, and let a be a fixed integer.
Suppose the following two statements are
true:
1. P(a) is true.
2. For all integers k ≥ a, if P(k) is true then
P(k+1) is true.
• Then the statement, for all integers n ≥ a,
P(a) is true
Example
• For all integers n ≥ 8, n cents can be obtained
using 3 cents and 5 cents.
• or, for all integers n ≥ 8, P(n) is true, where
P(n) is the sentence “n cents can be obtained
using 3 cents and 5 cents.”
Example
• This can be proven by
exhaustion if we can
continue to fill in the table
up to $1.00.
• The table shows how to
obtain k cents using 3 and 5
coins. We must show how to
obtain (k+1) cents.
• Two cases:
– k: 3 + 5, k+1: ?
• replace 5 with 3 + 3
– k: 3 + 3 + 3, k+1: ?
• replace 3+3+3 with 5+5
Method of Proof Mathematical
Induction
• Statement: “For all integers n≥a, a property P(n)
is true.”
• (basis step) Show that the property is true for n =
a.
• (inductive step) Show that for all integers k≥a, if
the property is true for n=k then it is true for
n=k+1.
– (inductive hypothesis) suppose that the property is
true for n=k, where k is any particular but arbitrarily
chosen integer with k≥a.
– then, show that the property is true for n = k+1
Example Coins Revisited
• Proposition 4.2.1: Let P(n) be the property “n
cents can be obtained using 3 and 5 cent
coins.” Then P(n) is true for all integers n≥8.
– Proof:
– Show that the property is true for n=8:
• The property is true b/c 8=3+5.
Example cont.
– Show that for all integers k≥8, if the property true
for n=k, then property true for n=k+1
• (inductive hypothesis) Suppose k cents can be obtained
using 3 and 5 cent coins for k≥8.
• Must show (k+1) cents can be obtained from 3 & 5 coin.
– Case (3,5 coin): k+1 can be obtained by replacing the 5 coin
with two 3 cent coins. This increments the value by 1 (3+3=6)
replaces the 5 cent coin.
– Case (3,3,3 coin): k+1 can be obtained by replacing the three 3
coins with two 5 coins. k=b+3+3+3=b+9 and
k+1=b+9+1=b+5+5=b+10
Example Formula
• Prove with mathematical induction
n(n 1)
1 2  ... n 
,n 1
2
• Identify P(n)
n(n 1)
P(n) 1 2  ... n 
,n 1
2

• Basis step
1(11) 2
P(1) 1 
 1
2
2


Example Formula cont.
• Inductive step
– assume P(k) is true, k>=1
k(k 1)
P(k) 1 2  ... k 
2 k+1 for n
– show that P(k+1) is true by subing
(k 1)(k 11) (k 1)(k  2)
P(k 1) 1 2  ... k 1 

2
2

– show that left side 1+2+…k+1 = right side (k+1)(k+2)/2
– 1+2+…+k+1 = (1+2+…+k) + k+1
– sub from inductive hypothesis: k(k 1)  (k 1)  k(k 1)  2(k 1)
2
(k 1)(k  2)

2
2
2
Theorem 4.2.2
• Sum of the First n Integers
– For all integers n≥1,
n(n 1)
1 2  ... n 
2

Example
• Sum of the First n Integers
– Find 2+4+6+…+500
•
•
•
•
Get in form of Theorem 4.2.2 (1+2+…+n)
factor out 2: 2(1+2+3+…+250)
sum = 2( n(n+1)/2 ), n = 250
sum = 2( 250(250+1)/2 ) = 62,750
– Find 5+6+7+8+…+50
• add first 4 terms 1+2+3+4 to problem then subtract back out
after computation with 4.2.2
• 1+2+3+4+5+6+7+…+50 – (1+2+3+4)
• (50 (50+1)/ 2) – 10
• =1,265
Sum of Geometric Sequence
r n 1 1
 r  r 1
i0
n
• Prove that
, for all integers n≥0 and all
real numbers r except 1.
r 1
r

• P(n):   r 1
• Basis:  r  r r 11  1

• Inductive:
(n=k)  r  r 1
r 1

• Inductive
Hypothesis: (n=k+1)
i
n 1
n
i
i0
0
01
i
i0
k 1
k
i
i0
k 1
r k 11 1 r k 2 1
 r  r 1  r 1
i0
i
k 1
k
r  r
i
i0
i0
i

r
k 1
r k 1 1 k 1 r k 1 1 r k 1(r 1) r k 1 1 r k 2  r k 1 r k 2 1

r 



r 1
r 1
r 1
r 1
r 1
Theorem 4.2.3
• Sum of Geometric Sequences
– For any real number r except 1, and any integer
n≥0,  r  r r 11
n 1
n
i
i0

Examples
• Sum of Geometric Sequences
– Find 1 + 3 + 32 + … + 3m-2
r n 1 1
 r  r 1
i0
n
i
• Sequence is in geometric series,
apply 4.2.3 directly

30  31  32  ... 3m 2
3m 21 1 3m 1 1


3 1
2
– Find 32 + 33 +… + 3m

• rearrange into proper geometric sequence by factoring
out 32 from sequence
m 2
m 1
3
1
2
i
• 32(1 + 3 + 32 + … + 3m-2) =3  3  3 1
i0
