The sum of the first eight Fibonacci numbers is equal to the tenth Fibonacci number minus one.
Exercise: Prove the generalized form of this claim.
Generalised (because I'm British) claim:
The sum of any 8 consecutive Fibonacci numbers n1 - n8 is equal to the Fibonacci number n10 - n2.
I could notate this more clearly if only I could be bothered to render it in LaTeX :P
Lets call the first in our set of consecutive numbers a
, and the second b
. Then we have the sequence:
n1 n2 n3 n4 n5 n6 n7 n8 n9 n10
a b a+b a+2b 2a+3b 3a+5b 5a+8b 8a+13b 13a+21b 21a+34b
The sum of the first 8 of these is 21a+33b.
The difference then between sum(n1:n8) and n10 is b
, which is exactly n2, and thus sum(n1:n8) is n10 - n2