Inspired by Jake Lai

First, observe the sum $\displaystyle \sum _{1 \leq i \leq j \leq 100 } r_{i}r_{j}$ has the condition $i \leq j$ . Thus, we can split it into

$\sum _{1 \leq i < j \leq 100 } r_{i}r_{j} + \sum _{k=1}^{100} r_{k}^{2}$

Let the coefficients be $a_{k} = (-1)^{k}F_{k+1}$ . We deal with the first sum using Vieta's formulas:

$\sum _{1 \leq i < j \leq 100 } r_{i}r_{j} = \frac{a_{2}}{a_{0}} = 2$

The second sum we resolve using Newton's identities:

$\sum _{k=1}^{100} r_{k}^{2} = a_{1} \left(\sum_{k=1}^{100} r_{i}\right)-2a_{2}$

$= 1(1)-2(2) = -3$

Summing the two, we get that

$\sum _{1 \leq i \leq j \leq 100 } r_{i}r_{j} = \sum _{1 \leq i < j \leq 100 } r_{i}r_{j} + \sum _{k=1}^{100} r_{k}^{2} = 2-3 = \boxed{-1}$

$\displaystyle \sum_{1\leq i \leq j \leq 100} r_{i}r_{j} = \sum_{k=1}^{100} r_{k}^{2} +\sum_{1\leq i < j \leq 100} r_{i}r_{j} = \left ( \sum_{k=1}^{100} r_{k} \right )^{2} - \sum_{1\leq i < j \leq 100} r_{i}r_{j}$

from vieta formula we know the answer is $(-(-1))^{2}-2 = \boxed{-1}$