As you work through the computation of the total energy in the final charge configuration, you will quickly appreciate the need for a compact notation, in this case the summation notation.

What is the potential when no charges are present? $V=0$, so no energy is needed to bring the first charge in.

What is the potential when the first charge, $q_1$, is brought in to position $\rr_1$? We have \begin{equation} V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{|\rr-\rr_1|} \end{equation} so the energy needed to bring in the second charge, $q_2$, to position $\rr_2$ is \begin{equation} q_2 V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\rr_2-\rr_1|} \end{equation}

Because of the superposition principle, the resulting potential is just the sum of the separate potentials due to each charge, and the energy needed to bring in the third charge is just the sum of two terms of the same form as above. Thus, using summation notation, the total energy needed to bring in all of the charges is given by \begin{equation} U = \sum_{i<j} \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{|\rr_j-\rr_i|} \end{equation}

We can express this result in a more useful form by noticing that the expression inside the summation is the same if $i$ and $j$ are interchanged. Thus, if we extend the sum to include $i>j$, we pick up the same terms again, resulting in an overall factor of 2. Factoring out the constant, we finally obtain \begin{equation} U = \frac12 \frac{1}{4\pi\epsilon_0} \sum_{i\ne j} \frac{q_i q_j}{|\rr_j-\rr_i|} \end{equation}