Mathematics Magazine for Grades 1-12
Let , (1)
be the recurrent sequence. Suppose that , , , , ,
Writing the terms of this sequence, we get
0, 1, 1, 2, 3, 5, 8, (2)
called the Fibonacci sequence and F is called the n-th Fibonacci numbers.
Fibonacci was one of the greatest European mathematician of the middle ages, his full name was Leonardo of Pisa, or Leonardo Pisano in Italian, he was born in Pisa (Italy), the city with the famous Leaning Tower, about 1175 AD.
The Lucas numbers are defined by the equations , (3)
satisfy the same recurrence
where, the first few are
1, 3, 4, 7, 11, 18, 29, 47, 76,123, ...
The French mathematician, Edouard Lucas (1842-1891), was the first who gave the series of numbers 0, 1, 1, 2, 3, 5, 8, 13, .. the name the Fibonacci Numbers.
Assuming that the sequence Fn has the form , where is a real parameter.
Substituting in (1)
But ("n N*), the last equality becomes
This is a quadratic equation with res pect to the real parameter l having the roots
Thus the sequences ,
verify the equality (1). So we conclude that the equation (1) can have more solutions. In general there are an infinite of the sequences verifying (1). Easy to observe that (1) has the form (6)
c2 are fixed real numbers, verifying (1), too.
Also you can prove that any sequence verifying (1) has the form (6).
For n = 0 and n = 1 in (6), we get the linear system
having the solutions
Finally, the general term of the Fibonacci sequence has the form
Some Proprieties of the Fibonacci sequence.
Summing all equalities we get ,
so (8) is
2. and 3. can be proven in a similar manner.
Proof: It is easy to observe that , .
From this we have successively the equalities:
Summing all these
equalities we get (9).
5. Prove that (10)
where Fn is the n-th term of the Fibonacci sequences.
Proof: Using (1) and (9) then (10) is easy to be proven.
But my goal is to prove (10) using the mathematical induction. Ill proceed by mathematical induction with respect .
For m = 1, the equality (10) becomes , this being evident, then
(10) is true for m=1. (For example when m = 2 the formula (10) is true
Assuming that (10) is true for m = k and m = k + 1.
Ill prove that (10) is true for m = k + 2, too.
Therefore, being true the equalities
equalities we get
in fact this is (10)
for m = k + 2.
6. Prove that (sometimes called double angle formula),
If in (10)
m = n
7. Prove that , n>1 (11)
Proof: Again Ill proceed for (11) by mathematical induction.
For n = 2 then (11) becomes , which is true. Thus (11) is true for n=2.
Assuming that (11) is true for n, Ill prove that it is true for n + 1, too.
So is true. Adding in both sides in the last equality the number
I get or
Next using (1) I have and finally .
Hence condition (11) is true for n + 1.