<s>[INST] 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? 
 [/INST] You can prove by induction that the sum of the squares of the first $6n$ terms is divisible by $8$.

First, show that this is true for $n=1$:
$\sum\limits_{m=1}^{6}F_m^2=40$
Second, assume that this is true for $n$:
$\sum\limits_{m=1}^{6n}F_m^2=8k$
Third, prove that this is true for $n+1$:
$\sum\limits_{m=1}^{6(n+1)}F_m^2=$
$\sum\limits_{m=1}^{6n+6}F_m^2=$
$\left(\color\red{\sum\limits_{m=1}^{6n}F_m^2}\right)+F_{6n+1}^2+F_{6n+2}^2+F_{6n+3}^2+F_{6n+4}^2+F_{6n+5}^2+F_{6n+6}^2=$
$\color\red{8k}+F_{6n+1}^2+F_{6n+2}^2+F_{6n+3}^2+F_{6n+4}^2+F_{6n+5}^2+F_{6n+6}^2=$
$\small8k+F_{6n+1}^2+F_{6n+2}^2+(F_{6n+1}+F_{6n+2})^2+(F_{6n+1}+2F_{6n+2})^2+(2F_{6n+1}+3F_{6n+2})^2+(3F_{6n+1}+5F_{6n+2})^2=$
$8k+16F_{6n+1}^2+40F_{6n+2}^2+48F_{6n+1}F_{6n+2}=$
$8(k+2F_{6n+1}^2+5F_{6n+2}^2+6F_{6n+1}F_{6n+2})$

Please note that the assumption is used only in the part marked red.
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