Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number? Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number? 

There is no need to find an explicit formula for $\sum_{n=1}^{600}F_n^2$ (but this is doable and I'll show it in the second part of this answer). The Fibonacci sequence obeys the relation $F_{n+2}=F_{n+1}+F_{n}$ and $8$ is a Fibonacci number, hence the sequence $\{F_n\pmod{8}\}_{n\geq 1}$ has a short period and the same applies to $\{F_n^2\pmod{8}\}_{n\geq 1}$:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}\hline  n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9 & 10 & 11 & 12\\ \hline F_n\pmod{8} & 1 & 1 & 2 & 3 & 5 & 0 & 5 & 5 & 2 & 7 & 1 & 0 \\ \hline F_n^2\pmod{8} & 1 & 1 & 4 & 1 & 1 & 0 & 1 & 1 & 4 & 1 & 1 & 0\\ \hline\end{array}$$
By direct inspection, we have that $F_{6n+1}^2+F_{6n+2}^2+F_{6n+3}^2+F_{6n+4}^2+F_{6n+5}^2+F_{6n+6}^2$ is always a multiple of $8$, hence $6\mid 600$ implies $8\mid\sum_{n=1}^{600}F_n^2$.

Now the brute-force approach. Let $\sigma=\frac{1+\sqrt{5}}{2}$ and $\bar{\sigma}=\frac{1-\sqrt{5}}{2}$. We have $\sigma+\bar{\sigma}=1$, $\sigma\bar{\sigma}=-1$ and
$$ F_n = \frac{\sigma^n-\bar{\sigma}^n}{\sqrt{5}},\qquad F_n^2 = \frac{\sigma^{2n}+\bar{\sigma}^{2n}-2(-1)^n}{5} $$
hence it is enough to show that $8$ is a divisor of $L_2+L_4+\ldots+L_{1200}$, where $L_n=\sigma^{n}+\bar{\sigma}^n$ is a Lucas number. Lucas numbers obey the same recurence relation as Fibonacci numbers, hence
$$ L_2+L_4+\ldots+L_{1200} = L_{1201}-L_{1} $$
can be easily proved by induction and it is enough to show that $L_{1201}\equiv L_1\pmod{8}$.
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline L_n\pmod{8} & 2 & 1 & 3 & 4 & 7 & 3 & 2 & 5 & 7 & 4 & 3 & 7 & 2 & 1 \\ \hline \end{array}$$
By direct inspection, the period of the sequence $\{L_n\pmod{8}\}_{n\geq 1}$ is $12$, hence $L_{1201}\equiv L_1\pmod{8}$ as wanted.
