Công thức: $1^2+2^2+3^2+...+n^2=\dfrac{n \left(n+1\right) \left(2n+1\right)}{6}$
Ta được:
$1^2+2^2+3^2+4^2+...+\left(2012^{2016}\right)^2=\dfrac{2012^{2016}.\left(2012^{2016}+1\right).\left(2.2012^{2016}+1\right)}{6}$
Ta có:
$2012^{2016} \equiv 2012^{2000}.2012^{16} \equiv 12^{2000}.12^{16} \equiv 376.416 \equiv 416$
$\Rightarrow 1^2+2^2+3^2+4^2+...+\left(2012^{2016}\right)^2 \equiv \dfrac{416.\left(416+1\right).\left(2.416+1\right)}{6} \equiv 696$
Vậy ba chữ số tận cùng của $1^2+2^2+3^2+4^2+...+\left(2012^{2016}\right)^2$ là 696.
