[Home]
## A Fifth GeneralizationWe derive some relationships of our recurrence formula
## 1. ∑ A ( n ) :We rewrite the equation for A ( n ): We sum up both sides of the equation over n yielding Observing A ( 2 ) = m · A ( 1 ) - A ( 0 ) finally yields The binomial summation formula is achieved using equation (30) Writing equation (30) in full yields Summing up over n yields We give an example for n = 5: m = 4 , A (1 ) = 1 , A ( 2 ) = 4 We present We ask now for a recurrence formula for ## 2. A ( 2n-1 ) , A ( 2n )We start from equation A ( n ) = m · A ( n-1 ) - A ( n-2 ). Observing yieldsSo we achieve a correlation between every second element
of the sequence which is of the same form as the starting
recurrence formula but m replaced by m ## 3. ∑ A ( 2n-1 ) , ∑ A ( 2n ) :We sum up both sides of the equations (63) and (64) over n and obtain We give an example for ∑ A ( 2n ) , m = -3 , seed ( - 2 , 1 ) and n = 5. We present ∑ A ( 2n ) in the left justified manner. We can derive formulae for Observing A ( 2 ) = m · A ( 1 ) - A ( 0 ) yields In a very similar way we can obtain In our next page we will present binomial summation formulae for the more general recurrence formula A ( n+2 ) = m · A ( n+1 ) + p · A ( n ) and G ( k n + l ) . You are kindly invited to visit this page. |