Answer
Calculate $A+(B+C)$ and $(A+B)+C$.
Work Step by Step
Consider the matrices with $a_{ij}$,$b_{ij}$ and $c_{ij}$ real numbers:
$$A=\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}\end{bmatrix}$$
$$B=\begin{bmatrix}b_{11}&b_{12}&...&b_{1n}\\b_{21}&b_{22}&...&b_{2n}\\...&...&...&...\\b_{m1}&b_{m2}&...&b_{mn}\end{bmatrix}$$
$$C=\begin{bmatrix}c_{11}&c_{12}&...&c_{1n}\\c_{21}&c_{22}&...&c_{2n}\\...&...&...&...\\c_{m1}&c_{m2}&...&c_{mn}\end{bmatrix}$$
First calculate $A+(B+C)$:
$$\begin{align*}
A+(B+C)&=\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}\end{bmatrix}+\left(\begin{bmatrix}b_{11}&b_{12}&...&b_{1n}\\b_{21}&b_{22}&...&b_{2n}\\...&...&...&...\\b_{m1}&b_{m2}&...&b_{mn}\end{bmatrix}+\begin{bmatrix}c_{11}&c_{12}&...&c_{1n}\\c_{21}&c_{22}&...&c_{2n}\\...&...&...&...\\c_{m1}&c_{m2}&...&c_{mn}\end{bmatrix}\right)\\
&=\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}\end{bmatrix}+\begin{bmatrix}b_{11}+c_{11}&b_{12}+c_{12}&...&b_{1n}+c_{1n}\\b_{21}+c_{21}&b_{22}+c_{22}&...&b_{2n}+c_{2n}\\...&...&...&...\\b_{m1}+c_{m1}&b_{m2}+c_{m2}&...&b_{mn}+c_{mn}\end{bmatrix}\\
&=\begin{bmatrix}a_{11}+(b_{11}+c_{11})&a_{12}+(b_{12}+c_{12})&...&a_{1n}+(b_{1n}+c_{1n})\\a_{21}+(b_{21}+c_{21})&a_{22}+(b_{22}+c_{22})&...&a_{2n}+(b_{2n}+c_{2n})\\...&...&...&...\\a_{m1}+(b_{m1}+c_{m1})&a_{m2}+(b_{m2}+c_{m2})&...&a_{mn}+(b_{mn}+c_{mn})\end{bmatrix}.
\end{align*}$$
Then calculate $(A+B)+C$:
$$\begin{align*}
(A+B)+C&=\left(\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}\end{bmatrix}+\begin{bmatrix}b_{11}&b_{12}&...&b_{1n}\\b_{21}&b_{22}&...&b_{2n}\\...&...&...&...\\b_{m1}&b_{m2}&...&b_{mn}\end{bmatrix}\right)+\begin{bmatrix}c_{11}&c_{12}&...&c_{1n}\\c_{21}&c_{22}&...&c_{2n}\\...&...&...&...\\c_{m1}&c_{m2}&...&c_{mn}\end{bmatrix}\\
&=\begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}&...&a_{1n}+b_{1n}\\a_{21}+b_{21}&a_{22}+b_{22}&...&a_{2n}+b_{2n}\\...&...&...&...\\a_{m1}+b_{m1}&a_{m2}+b_{m2}&...&a_{mn}+b_{mn}\end{bmatrix}+\begin{bmatrix}c_{11}&c_{12}&...&c_{1n}\\c_{21}&c_{22}&...&c_{2n}\\...&...&...&...\\c_{m1}&c_{m2}&...&c_{mn}\end{bmatrix}\\
&=\begin{bmatrix}(a_{11}+b_{11})+c_{11}&(a_{12}+b_{12})+c_{12}&...&(a_{1n}+(b_{1n})+c_{1n}\\(a_{21}+(b_{21})+c_{21}&(a_{22}+(b_{22})+c_{22}&...&(a_{2n}+(b_{2n})+c_{2n}\\...&...&...&...\\(a_{m1}+(b_{m1})+c_{m1}&(a_{m2}+(b_{m2})+c_{m2}&...&(a_{mn}+(b_{mn})+c_{mn})\end{bmatrix}.
\end{align*}$$
Because $a_{ij}$, $b_{ij}$ and $c_{ij}$ are real numbers and in the set of real numbers addition is associative , we have:
$$a_{ij}+(b_{ij}+c_{ij})=(a_{ij}+b_{ij}++c_{ij},\text{ where }i=1,2,...,m,j=1,2,...,n$$
We got that
$$A+(B+C)=(A+B)+C.$$
