Consider $$ \begin{equation*} \langle x(n)\rangle = \sum_{i}^{n} \Delta x_i = 0 \hspace{1cm} \Delta x_i=\pm l, \end{equation*} $$ since we have an equal probability of jumping either to the left or to right. The value of \( \langle x(n)^2\rangle \) is $$ \begin{equation*} \langle x(n)^2\rangle = \left(\sum_{i}^{n} \Delta x_i\right)\left(\sum_{j}^{n} \Delta x_j\right)=\sum_{i}^{n} \Delta x_i^2+ \sum_{i\ne j}^{n} \Delta x_i\Delta x_j=l^2n. \end{equation*} $$