A Third Generalization (Part 2)
1.
We start with F (n+2 ) = F (n+1 ) + F (n ) = F (n +1) + F ( n +1) - F (n-1 )
F (n+2 ) = 2 F ( n+1 ) - F ( n-1 )
Observing F ( n-1 ) = F ( n ) - F ( n-2 ) finally yields
F ( n+2 ) = 2 F ( n+1 ) - F ( n ) + F ( n-2 )
The binomial summation formula of the recurrence formula
with seed S ( A (1) , A(2) , A(3) , A(4) ) reads
With the seed S ( 1 , 1 , 2 , 3 ) we obtain the Fibonacci sequence and with the seed S ( 1 , 3 , 4 , 7 ) the Lucas sequence respectively :
We can rewrite the Fibonacci binomial formula in the more elegant form
Summing up over n yields
Below we give an example for ∑ F ( n ) of the first nine elements of the sequence.
Replacing n by 2n - 1 and 2n yields F ( 2n-1 ) and F (2n ) respectively:
2.
The generalized Fibonacci sequence can be expressed as follows :
Using the seed S ( 1 , 1 ) and S ( 1 , 3 ) yields the expressions for the Fibonacci and the Lucas sequence respectively :
Summing up over n yields the binomial summation formula for ∑ F ( n ) :
As an example again we present the first nine elements of the sequence
The binomial summation formulae for G ( 2n-1 ) and G ( 2n ) respectively read
Using the seeds S (1 , 1 ) for the Fibonacci sequence and S ( 1 , 3 ) for the Lucas sequence respectively yields
Summing up over n yields
As an example we present the first five elements of the sequence ∑ F ( 2n-1 ) :
3.
We use here another binomial sum formula for the generalised Fibonacci sequence of every second element G ( 2n-1 ):
With the seeds S ( 1 , 2 ) and S ( 1 , 4 ) we obtain the formulae for the Fibonacci and the Lucas sequences:
We give an example of the first five elements of the Fibonacci sequence:
4.
Another binomial summation formula for G ( 2n-1 ) reads
Again using the seed S ( 1, 2 ) yields F ( 2n-1 ) :
Below we give an example of the first nine elements of the sequence :
You are kindly invited to the next page where we will extend our binomial summation fomula
to the more general form
