Mathematical Induction
Name:
Mathematical induction is a method to prove that a statement about the positive integers
is true for every positive integer. Symbolically the process can be described as standing
an infinite set of dominos on end close to each other and then toppling the first domino
which in turn topples the next, which topples the next, and so on.
The Axiom of Mathematical Induction
Let P(n) be a statement about a positive integer n; let S be the set of all positive integers n
so that P(n) is true. Then S is the set of positive integers if and only if (a) 1 is in S and (b)
if k is in S, then k + 1 is in S.
n(n  1)
2
Now with the Axiom of Mathematical Induction, we can prove it is true.
Example 1: Earlier we conjectured that 1  2  ...  n 
1(1  1) 2
  1 [Note: this shows S is not the empty set.)
2
2
k (k  1)
(b) Select an arbitrary k so that 1  2  ...  k 
[Note: this is called the Inductive
2
Hypothesis. Since S is not empty, there is a k we can select.]
(k  1)((k  1)  1)
[Now we want to show that 1  2  ...  k  ( k  1) 
]
2
In the proof, we would not write this down at this time, but we would keep it in mind, or
write it down on a separate page so that we know where we want to go.
a) For n = 1,
At this stage there are two ways to proceed. The first is to start with the left side of our
conclusion: reduce the left side to the inductive hypothesis, then turn out attention to the
right side and complete the proof. In this case, the proof would go as:
k (k  1)
 (k  1) [From the inductive hypothesis]
2
k (k  1) 2( k  1)


[Find a common denominator]
2
2
1  2  ...  k  (k  1) 

k (k  1)  2(k  1)
2
[Add the fractions]
(k  1)( k  2)
[Factor]
2
(k  1)((k  1)  1)

[Write as desired.]
2
n(n  1)
Therefore, by Mathematical Induction, 1  2  ...  n 
for all n .
2
The alternate method is to use the Inductive Hypothesis and build up to the conclusion.

Mathematical Induction
Name:
For this problem, the proof would proceed as follows:
(b) Select and arbitrary k so that 1  2  ...  k 
1  2  ...  k  (k  1) 
k (k  1)
2
k (k  1)
 (k  1) [Add (k  1) to both sides]
2
Now the proof proceeds as above, the left side is as desired, so our attention can be
devoted to the right side.
[Note: the two methods appear similar, but the thought process is quite different. ]
Example 2:
Prove that the sum of the interior angles of a convex n-gon (a polygon with n-sides) is
180  ( n  2) .
Proof by Mathematical Induction
In this case, we will start with n = 3 instead of n = 1.
[It is proper to state the method of proof.]
[I think you know why.]
Step 1: Let n = 3 and we know that the sum of the measures of the angles of a triangle is
180  180  (3  2)  180  ( n  2) .
Step 2: Assume that for some k, the sum of the measure of the interior angles of the convex kgon is 180  ( k  2) . We need to show that it follows that the sum of the measures of the
interior angles of the convex (k + 1)-gon is 180  ((k  1)  2) .
To this end, let A1A2A3...AkAk+1 be a convex (k+1)gon. From our assumption that Sk is true, we know
that the sum of the angles of the polygon A1A2A3...Ak
is 180  ( k  2) . We need to consider how many
more degrees are added when the (k + 1) vertex is
added. From the figure at the right, it is clear that
the sum of the angles of A1A2A3...AkAk+1 = (the sum
of A1A2A3...Ak) + (the sum of AkAk+1A1)
= 180  ( k  2)  180
= 180  ( k  2  1)
= 180  (( k  1)  2)
which is what we wished to show.
A3
A2
Ak-1
Ak
A1
Ak+1
Therefore the sum of the interior angles of any convex n-gon is 180  ( n  2)
Now it is your turn to try some. Use Mathematical Induction to prove the following.
Remember to first show that the statement is true for a specific case (usually S1) and then
show the general case.
Mathematical Induction
Name:
Problem 1:
Prove proposition Pn: 1 + 2 + 22 + … + 2n = 2n+1 – 1 for integers n  0.
Step 1: Show that this is true for n = 0.
Step 2: Assume that Pk is true:
Pk:
[Show the finished statement for Pk+1 to know where you're headed.]
We wish to prove:
Pk+1:
Proof:
Which is what we wished to derive.
Therefore,
Mathematical Induction
Name:
Induction Problems. Prove each of the following statements using Mathematical
Induction.
n
1) For all positive integers n, (i.e. n
),
 k (k  1)  n  1
1
n
k 1
n
2) For all n
,
k
k 1
2

n  (n  1)(2n  1)
6
n
3) For all n
,
 k  k !  (n  1)! 1
k 1
n
4) For all n
,
 k  (k  1) 
k 1
n  (n  1)  (n  2)
3
5) For all n
, 35 divides 62n  1
6) For all n
, 24 divides 52n  1
7) For all n 
with n  2 , 2n  2n  1
8) For all n 
with n  2 , The number of diagonals in a convex n-gon is
9) For all n 
same size).
with n  5 , a square can be divided into n squares (not necessarily the
n  (n  3)
2