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Let $I,J,K$ be ideals in ring $R$, then prove that $(IJ)K=I(JK)$.

Take an element from $(IJ)K$, then the element is the finite sum of $ab$ where $a\in IJ$, $b \in K$. Then since $a\in IJ$, $a=\sum_{p=1}^{n} i_p j_p$. Then $ab=(\sum_{p=1}^n i_pj_p)b$. I don't know how to make this summation in the form of $c(\sum_{p=1}^m j_p k_p)$ where $c \in I$.

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    This is easy: both ideals are generated by $\{ ijk : i \in I , j \in J , k \in K\}$. Since they have the same set of generators, they are equal. – 2017-02-03

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$$\sum_{q = 1}^m \left(\sum_{p=1}^n i_{p,q} j_{p,q} \right) k_q = \sum_{p,q} i_{p,q}j_{p,q}k_q = \sum_{p,q} i_{p,q} (j_{p,q}k_q)$$