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

Proof:

Lets call the first in our set of consecutive numbers

a, and the second

b. Then we have the sequence:

**Code:**

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