No Fibbing 3

$\begin{aligned} \sum_{k=1}^n F_k F_{k+1} F_{k+2} &= \sum_{k=1}^n \left(F_{k+2} - F_{k+1} \right) F_{k+1} F_{k+2} \\ &= \sum_{k=1}^n \left(F_{k+1} F_{k+2}^2 - F_{k+1}^2 F_{k+2} \right) \\ &= \sum_{k=1}^n \left(F_{k+1} F_{k+2}^2 - F_{k+1}^2 \left(F_k + F_{k+1} \right) \right) \\ &= \sum_{k=1}^n \left(F_{k+1} F_{k+2}^2 - F_kF_{k+1}^2 - F_{k+1}^3 \right) \\ & = F_{n+1}F_{n+2}^2 - F_1F_2^2- \sum_{k=1}^n F_{k+1}^3 \\ & = F_{n+1}F_{n+2}^2 - 1 - \sum_{k=2}^{n+1} F_k^3 \\ & = F_{n+1}F_{n+2}^2 - \sum_{k=1}^{n+1} F_k^3 \end{aligned}$

Therefore,

$\begin{aligned} Q & = \frac {\sum_{k=1}^n F_k^3 + \sum_{k=3}^n F_k F_{k-1} F_{k-2}}{\sum_{k=1}^n F_k^2} \\ & = \frac {\sum_{k=1}^n F_k^3 + \color{#3D99F6} \sum_{k=1}^{n-2} F_k F_{k+1} F_{k+2}}{\color{#D61F06}\sum_{k=1}^n F_k^2} & \small \color{#D61F06} \text{Note that }\sum_{k=1}^n F_k^2 = F_n F_{n+1} \\ & = \frac {\sum_{k=1}^n F_k^3 + \color{#3D99F6} F_{n-1}F_n^2 - \sum_{k=1}^{n-1} F_k^3}{\color{#D61F06} F_n F_{n+1}} \\ & = \frac {F_n^3 + F_{n-1}F_n^2}{F_n F_{n+1}} = \frac {F_n^2(F_{n-1}+F_n)}{F_n F_{n+1}} \\ & = \frac {F_n^2F_{n+1}}{F_n F_{n+1}} = F_n \end{aligned}$

For $n=100$ , the answer is $\boxed{F_{100}}$ .

It can be proved geometrically that ${\displaystyle \sum_{n=1}^{x} ({F_{n}} ^3)} = (F_x)^2{F_{x+1}} - \displaystyle \sum_{n=3}^{x} F_n F_{n-1} F_{n-2}$ for all $x \geq 3$

Rearranging we get $\displaystyle \sum_{n=1}^{x} ({F_{n}} ^3) + \displaystyle \sum_{n=3}^{x} F_n F_{n-1} F_{n-2} = (F_x)^2 {F_{x+1}}$

It can also be shown geometrically that $F_x F_{x+1} = \displaystyle \sum_{n=1}^{x} ({F_{n}} ^2)$

Therefore $\displaystyle \sum_{n=1}^{x} ({F_{n}} ^3) + \displaystyle \sum_{n=3}^{x} F_n F_{n-1} F_{n-2} = (F_x) \displaystyle \sum_{n=1}^{x} ({F_{n}} ^2)$

Rearranging we get $\frac {\displaystyle \sum_{n=1}^{x} ({F_{n}} ^3) + \displaystyle \sum_{n=3}^{x} F_n F_{n-1} F_{n-2}}{\displaystyle \sum_{n=1}^{x} ({F_{n}} ^2)}$ $=$ $F_x$

When $x = 100$ the equation is equal to $\boxed{F_{100}}$