Sequences and Series
Section 12-9
Mathematical Induction
Mathematical Induction

It is a method of proof.

It depends on a recursive process that works much like an unending line of
dominos arranged so that if any one domino falls the next one will also fall.

Mathematical Induction:

S1 is true condition 1

S1 implies S2 is true condition 2

S2 implies S3 is true condition 3

And so forth
Proof by Mathematical Induction
 First, verify
that the conjecture Sn is valid
for the first possible case, usually n=1.
This is called the anchor step.
 Then
assume that Sn is valid for n=k, and
use this assumption to prove that it is
also valid for n=k+1. This is called the
induction step.
Example 1

Prove that the sum of the first n even positive integers is n (n+1).
Step # 1 verify that it is valid for n=1
1(1+1) = 2
Step # 2 Assume that the formula is valid for n=k and prove that it is also valid for n=k+1
Sk
2+ 4+ + 6+ … + 2k = k(k+ 1)
Sk+1
2+ 4+ 6+ … .+ 2K + 2(k+ 1)= K (K + 1) +
2(K+1)= (k+2)(K+1).
If k+1 is substituted into the original formula
The same result is obtained . Thus if the formula IS VALID FOR N=K, IT IS ALSO VALID FOR
N = k+ 1.Since n= 1 ho ld s it is valid fo r n= 2… .n= 3 and so o n.
Example # 2
n
 Prove 6 -1 is divisible by 5 for all positive integers.

Step # 1:Verify that it holds for n=1 Since 5 is divisible by 5 , it
holds.

Assume that 6 -1 is divisible by 5 for n=k and use this assumption
to prove that it is also divisible by 5 for n=k+1

k
Sk
 Sk+1
k
6 -1 =5r
6
k 1
6 -1 = 5r for some integer r then
k 1
6 -1 = 5t for some integer t
k
6 (6 -1) = 6(5r)
6
k 1
-6 = 30r
-1 = 5(6r+1) Let t = 6r+1 an integer. Then 6 k  1 -1 = 5t. We have
shown that if it is valid for Sk then it is valid for sk+1. Since it is valid for n=1, it is
valid for n=2 and so on. Thus 6
n
-1 is divisible by 5 for all positive integers n.
HW#45
Section 12-9
Pp. 826-828
#11,13,17,23,29,30