An infinite number of charges, each equal to q coulomb, are placed along the x-axis at x (in metres)= 1, 2, 4, 8, . and so on. If the potential  due to charges are given by $\frac{1}{4\pi {\epsilon }_0}\underset{ }{\sum _{i=1}^n\frac{q_i}{r_i} }$then potential in SI units at x = 0. due to this set of charges is$\frac{1}{4\pi {\epsilon }_0}$ times

Given that infinite number of charges, each equal to q coulomb, are placed along the x-axis at x (in metres)= 1, 2, 4, 8, . and so on. 

The potential and field in SI units at x = 0. 

due to this set of charge are respectively $\frac{1}{4\pi {\epsilon }_0}$ times

An infinite number of charges, each equal to q coulomb, are placed along the x-axis at x (in metres)= 1, 2, 4, 8, . and so on. The potential and field in SI units at x = 0. due to this set of charge are respectively $\frac{1}{4\pi {\epsilon }_0}$ times =$2\mathrm{q},\frac{4\mathrm{q}}{3}$.

$\begin{array}{l}V=\frac{kq}{1}+\frac{kq}{2}+\frac{kq}{4}+\frac{kq}{8}+\dots .\\ V=kq\left[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots .\right]\\ V=kq\left[\frac{1}{1-\frac{1}{2}}\right]\\ V=2kq\end{array}$

Here, 

$k=\frac{1}{4\pi {\epsilon }_0}$

So,

$V=2q\frac{1}{4\pi {\epsilon }_0}$

So, we get charge 2q times $\frac{1}{4\pi {\epsilon }_0}$.