G/ Accelerated Math 3
Mathematical Induction (9.4)
Name: ______________________
Date: _______________________
Standard: MA3A9. Students will use sequences and series
d. Use mathematical induction to find and prove formulae for sums of finite series.
Mathematical induction is a method of mathematical proof typically used to establish that
a given statement is true of all natural numbers (ordinary counting
numbers 1, 2, 3, ...). It is done by proving that the first statement in the
infinite sequence of statements is true, and then proving that if any one
statement in the infinite sequence of statements is true, then so is the next
one.
An informal description of mathematical induction can be illustrated by
reference to the sequential effect of falling dominoes.

---------------------------------------------------------------------------------------------------------------The Principle of Mathematical Induction
Let Sn be a statement involving the positive integer n. If
1. Sn is true, and
2. for every positive integer k, the truth of Sk , implies the truth of Sk 1 then the
statement Sn must be true for all positive integers n.
---------------------------------------------------------------------------------------------------------------
Prove the following are true for every positive integer n using mathematical


induction.

Example 1: 3  7  11  15   (4n  1)  n(2n  1) .
Helpful hints & explanations:
Show that Sn is true for n=1.
Sn  n(2n 1)  S1 1(2(1) 1)  3
2a. Assume that Sn is true for Sk , where k is equal to
any integer.
S
 (4k  1)  k (2k  1)
k  3  7  11  15 



b. Show that Sk 1 is true.


Sk1  (k 1)(2(k 1) 1)
1.
3  7  11  15 

 (4k  1)  (4(k  1) -1)  k (2k  1)  (4( k  1)  1)
 2k 2  k  (4k  4 1)

 2k 2  5k  3
 2k 2  3k  2k  3
 k(2k  3)  1(2k  3)

 (k  1)(2k  3)
 (k  1)(2k  2  1)
 (k  1)(2(k  1)  1)
3. Concluding statement:

Always write down for yourself
what Sk 1 equals so that you know
what you are trying to achieve.
At step 2b you are adding one more
term to the equation, k+1. What you
add to one side of the equation must
be added to the other side of the
equation so that the equation
remains balanced.
The RIGHT side of the equation is
simplified to show that it is equal to
Sk 1 .
The end of every proof always
contains a concluding statement
stating what has been proven.
Thus, Sk 1 is true. By the Principle of Mathematical Induction Sn is true for every positive
integers n.
Example 2: 5  7  9  11  13   (3  2n)  n(n  4) .
 1.

Show that Sn is true for n=1.
2a. Assume that Sn is true for Sk , where k is equal to any integer.

b. Show that Sk 1 is true.



Sk1 

3. Concluding statement:
---------------------------------------------------------------------------------------------------------------Example 3: 1  2  22  23   2n1  2n  1 .
1.
Show that Sn is true for n=1.
2a. Assume that Sn is true for Sk , where k is equal to any integer.

b. Show that Sk 1 is true.



3. Concluding statement:
Sk1 
