The Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13...; starting with 0 and 1, each number is the sum of the two previous numbers.
The Lucas numbers follow the same rule, but start with 2 and 1: 2, 1, 3, 4, 7, 11, 18,...
What's the sum of the first k Lucas numbers?
What's the sum of the squares of the first k Lucas numbers?
Since the same recurrence is satisfied, one expects the same, or very similar, results for the Lucas numbers as for the Fibonacci numbers which we treated in "Fibonacci sums" by Federico Kerecki. In fact the sum of the Lucas numbers from 0 to N has the same formula as for the Fibonacci numbers: the N+2nd one less 1. This should be no surprise, especially in light of the well-known formula L(n)=F(n+1)-F(n-1). For the sum of the squares of the Lucas numbers from 0 to N, I see L(N)xL(N+1)+2 by looking at Charlie's numbers.
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Posted by Richard
on 2004-10-07 14:11:29 |