Successively longer sums of consecutive Fibonacci numbers: pattern? Consider the following:
$$\begin{align}
F_{n-1}+F_{n-2}&=F_n\\
F_{n-1}+F_{n-2}+F_{n-3}&=F_{n-1}+F_{n-1}\\
&=2F_{n-1}\\
F_{n-1}+F_{n-2}+F_{n-3}+F_{n-4}&=F_n+F_{n-2}\\
&=L_{n-1}\\
F_{n-1}+F_{n-2}+F_{n-3}+F_{n-4}+F_{n-5}&=F_{n-1}+L_{n-2}\\
&=F_{n-1}+F_{n-1}+F_{n-3}\\
&=2F_{n-1}+F_{n-3}\\
F_{n-1}+F_{n-2}+F_{n-3}+F_{n-4}+F_{n-5}+F_{n-6}&=2(F_{n-1}+F_{n-4})
\end{align}$$
Is there some logic to this pattern? Can it be predicted? Is there a formula that can be used to compute the value of
$$\sum_{i=1}^{r}{F_{n-i}}$$
in terms of only Fibonacci and Lucas numbers or their multiples, no constant terms, without resorting to a summation?
 A: $$F_n = \left(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n\right)_{2,1}$$
which let's us treat it like any other summation except using matrices instead of scalar numbers, just remember that matrix multiplication doesn't commute and everything else is the same.
Using $X = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, proceed:
$$\begin{align}
S &= \sum_{i=1}^r F_{n - i} \\
  &= \left(\sum_{i=1}^r X^{n-i}\right)_{2,1}\\
  &= \left(\sum_{i=n - r}^{n - 1} X^i\right)_{2,1} \\
  &= \left(X^{n - r}
     \underbrace{\sum_{i=0}^{r - 1} X^i}
     _\text{Q = Geometric Series}
     \right)_{2,1}\\
\end{align}$$
Geometric series,
$$\begin{align}
Q &= \sum_{i=0}^{r - 1} X^i\\
  &= \left(X^{r} - I\right)
     \left(X     - I\right)^{-1}\\
  &= \left(X^{r} - I\right)
     X\\
  &= X^{r+1} - X
\end{align}$$
$$\begin{align}
S &= \left(X^{n - r}\right)
     \left(X^{r+1} - X\right)_{2,1}\\
  &= \left(X^{n+1} - X^{n+1-r}\right)_{2,1}\\
  &= F_{n + 1} - F_{n + 1 - r}
\end{align}$$
A: Using the formula that I found in this question, I have derived the following:
$$\begin{align}
f(n,r)&=\sum_{i=1}^r{F_{n-i}}\\
&=\sum_{i=1}^{n-1}{F_i}-\sum_{i=1}^{n-r-1}{F_i}\\
&=F_{n-1+2}-1-F_{n-r-1+2}+1\\
&=F_{n+1}-F_{n-r+1}\\
\\
f(8,6)&=F_{8+1}-F_{8-6+1}\\
&=F_9-F_3\\
&=34-2\\
&=32\\
&=13+8+5+3+2+1\\
&=F_7+F_6+F_5+F_4+F_3+F_2\\
\\
f(7,3)&=F_8-F_5\\
&=21-5\\
&=16\\
&=8+5+3\\
&=F_6+F_5+F_4
\end{align}$$
So it turns out the solution is the difference of two Fibonacci numbers.
