# Induction Solutions ```Mathematical Investigations IV
Name:
Mathematical Induction - Let the Dominoes Fall
Prove each statement using mathematical induction.
1.
Prove that for all positive integers, n,
1
1
1
n
=

 ... 
2n  1
1 3 3  5
(2n  1)(2n  1)
Proof(by induction):
Base Case: If n = 1, then LHS =
1
1
1
1
 and RHS =

1 3 3
2 1  1 3
Inductive Step: suppose that for some value of n,
1
1
1
n
=

 ... 
2n  1
1 3 3  5
(2n  1)(2n  1)
Then,
1
1
1
1
n
1

 ... 



1 3 3  5
(2n  1)(2n  1) (2n  1)(2n  3) 2n  1 (2n  1)(2n  3)
n(2n  3)
1


(2n  1)(2n  3) (2n  1)(2n  3)
2n 2  3n  1
(2n  1)(2n  3)
(2n  1)(n  1)

(2n  1)(2n  3)
(n  1)

(2(n  1)  1)

So for every natural number
1
1
1
n
=

 ... 
2n  1
1 3 3  5
(2n  1)(2n  1)
Mathematical Investigations IV
2.
Name:
Prove that for all positive integers, n,
1 3  2  4  3  5 
 n  ( n  2) 
n(n  1)(2n  7)
6
Proof (by induction):
Base Case (n = 1) If n = 1, LHS = 1 3  3 and RHS =
n(n  1)(2n  7) 1  2  9

 3.
6
6
Inductive Step:
Suppose that for some integer n, 1  3  2  4  3  5 
 n  ( n  2) 
n(n  1)(2n  7)
.
6
Then,
1 3  2  4  3  5 
 n  (n  2)  (n  1)(n  3) 
n(n  1)(2n  7)
 (n  1)(n  3)
6
 n(2n  7)

 (n  1) 
 (n  3) 
6


 2n 2  7n 6n  18 
 (n  1) 

6
6 

 2n 2  13n  18 
 (n  1) 

6


 (2n  9)( n  4) 
 (n  1) 

6

(n  1)(n  2)(2(n  1)  7)

6
So, 1  3  2  4  3  5 
 n  ( n  2) 
n(n  1)(2n  7)
holds for all positive integers n.
6
Mathematical Investigations IV
3.
Prove that
Name:
22n1  1 is divisible by 3 for all positive integers, n.
Proof by Induction:
Base Case (n = 1) If n = 1, then 22 n1  1  23  1  9 , which is divisible by 3. Therefor the
theorem holds when n = 1.
Inductive Step: Suppose that 22k 1  1 is divisible by 3 for some positive integer, k. Then
22 k 1  1  3m for some positive integer m. It follows that 22 k 1  3m  1
Then,
22( k 1) 1  1  22 k 3  1
 22 k 1  22  1
 4  22 k 1  1
 4(3m  1)  1
 12m  4  1
 12m  3
 3  (4m  1)
Which is clearly a multiple of 3.
Therefore, by PMI, 22n1  1 is divisible by 3 for all positive integers, n.
Mathematical Investigations IV
4.
Prove that
Name:
n
1

for all positive integers, n.
n 1 2
Proof (by induction):
Base Case (n = 1) If n = 1, LHS =
Inductive Step: Suppose that
Then
1
1
1
 which is clearly larger or equal to .
11 2
2
k
1
 for some positive integer k.
k 1 2
k 1
k 1
.

(k  1)  1 k  2
k 1
k
k
1

 .
, we will be done, since
k  2 k 1
k 1 2
To that end, not that for positive integer k:
If we can show that
k 1
k
(k  1)(k  1)
k (k  2)



k  2 k  1 (k  2)(k  1) (k  1)(k  2)
k 2  2k  1  ( k 2  2k )
(k  2)(k  1)
1

(k  2)(k  1)
0

k 1
k
k 1
k
1

 0, it follows that

 .
k  2 k 1
k  2 k 1 2
n
1

So by PMI,
for all positive integers, n.
n 1 2
Since
Mathematical Investigations IV
Name:
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