A sequence of Fibonacci facts

The Hampshire College Summer Studies in Mathematics program has what it calls an “Interesting Test” that interested students are invited to take as part of their application. Their web page offers a sample of past test problems. One of these, under the heading “Problems which at first seem to lack sufficient information”, is as follows:

From the third term on, each term of the sequence of real numbers $a_1, a_2, \ldots, a_{10}$ is the sum of the preceding two terms; that is, $a_n = a_{n-1} + a_{n-2}$ for $n = 3, 4, 5, \ldots, 10$. If $a_7 = 17$, what is the sum of all the ten terms?

I’d heard this one before (though it’s hard to find a source for things like this – the cloesst I could find was this Math StackExchange question. The solution is as follows: we can write $a_3, a_4, \ldots, a_{10}$ as linear combinations of $a_1, a_2$. In particular we have $a_3 = a_1 + a_2, a_4 = a_1 + 2a_2, \ldots, a_{10} = 21a_1 + 34 a_2$, where the general term is $a_n = F_{n-2} a_1 + F_{n-1} a_n$ (we’ll use this later). Here $F_n$ is the $n$th Fibonacci number, where $F_1 = F_2 = 1, F_n = F_{n-1} + F_{n-2}$. Adding everything up we get

$a_1 + \ldots + a_{10} = \left( 1 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21\right) a_1 + \left( 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 \right) a_2$

and doing those sums gives $a_1 + \ldots + a_{10} = 55 a_1 + 88 a_2$. But $a_7 = 5 a_1 + 8 a_2$, and so the sum is just $11a_7 = 11 \cdot 17 = 187$. For example, consider the sequence with $a_1 = 5, a_2 = -1$:

$5, -1, 4, 3, 7, 10, 17, 27, 44, 71$

which you can easily verify sums to 187. In the case where you actually have the whole sequence, this is a reasonably well-known “arithmetic trick” that one can show grade school children! The HCSSiM version, of course, doesn’t let you see the whole sequence.

But are there similar tricks to this with shorter or longer sequences? That’s usually the way with the Fibonacci numbers; facts about them tend to be parametrizable. A bit of experimentation shows us that with six terms we have

$a_1 + \ldots + a_6 = 8 a_1 + 12 a_2 = 4 (2a_1 + 3a_2) = 4 a_5.$

See for example the series $4, 1, 5, 9, 14, 23$; its sum, 56, is four times its fifth term. And with fourteen terms we have

$a_1 + a_2 + \ldots + a_{14} = 377 a_1 + 609 a_2 = 29 (13a_1 + 21a_2) = 29 a_9$

(I’ll leave you to come up with your own example.) It would appear that we have $a_1 + \ldots + a_{4k+2} = L_{2k+1} a_{2k+3}$, where the $L_j$ are the Lucas numbers $L_1 = 1, L_2 = 3, L_n = L_{n-1} + L_{n-2}$. Consider the sum of the first $4k+2$ terms satisfying a Fibonacci recurrence:

$a_1 + a_2 + \ldots + a_{4k+2} = F_{4k+2} a_1 + (F_{4k+3}-1) a_2.$

Now we have $F_{4k+2} = F_{2k+1} L_{2k+1}$ and $F_{4k+3}-1 = F_{2k+2} L_{2k+1}$. These are both special cases of the identity givnen in Wikipedia, $F_{n+k} + (-1)^k F_{n-k}= L_k F_n$. So we have

$a_1 + a_2 + \ldots + a_{4k+2} = F_{2k+1} L_{2k+1} a_1 + F_{2k+2} L_{2k+1} a_2 = L_{2k+1} (F_{2k+1} a_1 + F_{2k+2} a_2) = L_{2k+1} a_{2k+3}$.

So the sum of a fourteen-term Fibonacci sequence is equal to a multiple of the ninth term; the sum of a ten-term Fibonacci sequence, a multiple of the seventh term; the sum of a six-term Fibonacci sequence, a multiple of the fifth term. What of a sum of a two-term Fibonacci sequence? We get $a_1 + a_2 = L_1 a_3$. Recall $L_1 = 1$ — the sum of the first two terms is, trivially, a multiple of the third term.

(I don’t claim this is original. Only that I had some fun thinking about it. That probably means I need to get a life.)