We have
$$ F_{n+m}=F_nF_{m+1}+F_{n-1}F_{m}$$
Indeed, or $m=0$ the claim is $F_n=F_n\cdot 1+F_{n-1}\cdot 0$; for $m=1$ the claim is $F_{n+1}=F_n\cdot 1+F_{n-1}\cdot 1$; and for larger $m$ the claim follows from adding the corresponding equalities for $m-1$ and $m-2$.
Define $$P(i,k):=F_{i+1}\cdot F_{i+2}\cdot\ldots\cdot F_{i+k}.$$ Then we have the identity
$$ \begin{align}P(i,k)&=P(i,k-1)F_{i+k}\\&=P(i,k-1)(F_kF_{i+1}+F_{k-1}F_i)\\&=P(i,k-1)F_kF_{i+1}+P(i-1,k)F_{k-1}\end{align}$$
This allows us to show the
Claim. For $k\ge1$, $i\ge1$ we have $P(1,k)\mid P(i,k)$.
Proof. The case $i=1$ is trivial, as i sthe case $k=1$.
For all other cases, we use the above identity: If $P(i,k-1)=cP(1,k-1)$ and $P(i-1,k)=dP(1,k)$, then
$$P(i,k)=P(i,k-1)F_kF_{i+1}+P(i-1,k)F_{k-1}=(cF_{i+1}+dF_{k-1})P(1,k)$$
as was to be shown. $_\square$
Just write every term in the sum in terms of $a_1$ and $a_2$ (keeping in mind that $a_{n+2}=a_n+a_{n+1}$):
$$a_1+a_2+(a_1+a_2)+(a_1 + 2a_2)+(2a_1+3a_2)+(3a_1+5a_2)+(5a_1+8a_2)+(8a_1+13a_2)+(13a_1+21a_2)+(21a_1+34a_2). $$
Then the sum is clearly equal to $55a_1+88a_2 = 11(5a_1+8a_2)$, which is $11$ times the seventh term of the sum.
Best Answer
You want to find the $\gcd$ of the numbers
$$F_1+\cdots+F_n,\quad F_2+\cdots+F_{n+1},\quad F_3+\cdots+F_{n+2},\quad F_4+\cdots+F_{n+3},\quad \cdots\cdots$$
Since $F_1+\cdots+F_n=F_{n+2}-1 \implies F_{1+k}+\cdots+F_{n+k}=F_{n+k+2}-F_{k+2}$, this is
$$F_{n+2}-F_2,\quad F_{n+3}-F_3,\quad F_{n+4}-F_4,\quad F_{n+5}-F_5,\quad \cdots$$
Notice how this is a new Fibonacci(-like) sequence, in which each term is a sum of the two previous terms. Therefore, not only does the $\gcd$ of the whole sequence divide the $\gcd$ of the first two terms (this holds generically for any sequence), but the $\gcd$ of the first two terms divides all of the other terms hence divides the $\gcd$ of the whole sequence. Therefore, the $\gcd$ of the whole sequence is equal to the $\gcd$ of the first two terms! Using the algorithm $(a,b)=(a,b-a)$, we can simplify: $(F_{n+2}-F_2,F_{n+3}-F_3)=(F_{n+2}-F_2,F_{n+1}-F_1)=(F_n-F_0,F_{n+1}-F_1)$.
Therefore, the generalization of $3\mid(F_{k+1}+\cdots+F_{k+8})$ for all $k\ge0$ is
$$\gcd(F_n,F_{n+1}-1)\mid(F_{k+1}+\cdots+F_{k+n}).$$
Note this holds for negative values of $k$ too.