Download MEI Maths Item of the Month April 2015 Consecutive Fibonacci Squares

MEI Maths Item of the Month
April 2015
Consecutive Fibonacci Squares
The first few Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34 ... (each number is the sum of
the previous two numbers in the sequence and the first two numbers are both 1).
The sums of the squares of some consecutive Fibonacci numbers are given below:
12 + 12 = 2
32 + 52 = 34
132 + 212 = 610
Is the sum of the squares of consecutive Fibonacci numbers always a Fibonacci number?
Solution
The nth term of the Fibonacci sequence is:
n
 1 5   1 5 

 

2
2 



Fn 
5
n
The squares of two consecutive terms can be written as:
  1  5 n  1  5 n 

 
 
 2   2  

Fn 2  
5
2n
2
n
n
 1 5 
 1 5  1 5  1 5 

  2
 
 

2 
2   2   2 



5
n 1
n 1 2
  1 5 
1 5  

 
 
 2 
2  


Fn 12  
5
 1 5 


2 


2n2
1 5 
 2

 2 
n 1
1 5 


 2 
n 1
2n
1 5 


 2 
2n2
5
2
 1 5   1 5 

 

2   2 


2n2
2n
n
n
2
 1  5  1  5  1  5   1  5   1  5   1  5 
 2


 
 
 

 2  2  2   2   2   2 
5
n
n
 3  5  1  5 
 1  5   1  5   3  5  1  5 


  2
 
 


2  2 
2   2   2  2 



5
1 of 2
2n
2n
TB v1.0 © MEI
30/07/2015
MEI Maths Item of the Month
Adding the squares of two consecutive terms gives:
Fn 2  Fn 12
 1 5 


2 


2n
 1 5 
 

 2 
2n
 5  5   1 5 
 
 

2
2



2n
 3  5   1 5 
 
 

 2  2 
2n
2 n 1
2n
5
 1 5   1 5 
5
 2   2 




 1 5 


2 


 3  5   1 5 
 
 

 2  2 
5
 5  5   1 5 

 

2
2



 1 5 


2 


2n
2n
 1 5 

 2 
5
 1 5   1 5 
 5
 2   2 



2n
5
2 n 1
 
2 n 1
 1 5 
 

 2 
5
2 n 1
 F2 n 1
Which is a Fibonacci number.
2 of 2
TB v1.0 © MEI
30/07/2015