Quiz—Induction
Name: ______________
No Calculator.
1) Using the Principle of Mathematical Induction, prove that
1 1
1
1
for all integers n  0 , 1   2   n  2  n .
2 2
2
2
Proof (by Induction):
Base Case: n = 0
LHS = 1 and RHS = 2 
1
 2  1  1 . So base case is valid.
20
Inductive step: Assume for some k  0 , 1 
1 1


2 22
Then,
 1 1
1  2  22 


This completes the proof.
1
1
1
1

 k 1   2  k   k 1
k 
2  2
2  2

1
1
 2  k  k 1
2
2
2
1
 2  k 1  k 1
2
2
1 2
 2  k 1
2
1
 2  k 1
2

1
1
 2 k .
k
2
2
2)
Prove one of the following using Mathematical Induction. Clearly mark which one you are
doing for credit.
a. For all n
, 8 divides 32n – 1.
1 if n  1

b. Let an  
. Prove that for all n
1  2an 1 if n  2
, an  2 n  1
.
a. Proof (by Induction):
Base Case (n = 1) 32 – 1 = 9 – 1 = 8 is clearly divisible by 8.
Inductive step: Suppose that for some k 32 k  1 is divisible by 8. Then there is an integer
a such that 32 k  1  8a or 32 k  8a  1 .
Then,
32( k 1)  1  32 k  2  1
 3232 k  1
 25  32 k  1
 25  (8a  1)  1
 25  8a  25  1
 25  8a  24
 8(25a  3)
Since this last expression is divisible by 8, we have shown 32 n  1 divisible by all n
b. Proof (by Induction):
1 if n  1

 a1  1. Also, 21  1  1
Base Case (n = 1) an  
1

2
a
if
n

2
n 1

Inductive step: Suppose that for some k ak  2k  1 . Then,
ak 1  1  2 ak
 1  2(2 k  1)
 1  2 k 1  2
 2 k 1  1
So by induction, for all n
, an  2 n  1