Identical point masses each equal to m are placed at $x=0,x=1,x=2,x=\mathrm{4.......}$ The total gravitational force on mass m at $x=0$ due to all other masses is

the net force on particle x = o due to all other particles  

$\text{F}={\text{F}}_{\text{1}} +{\text{ F}}_{\text{2}} +{\text{ F}}_{\text{3}} +\mathrm{........}$

$F=\frac{{\text{Gm}}^{\text{2}}}{{\text{1}}^{\text{2}}}+\frac{{\text{Gm}}^{\text{2}}}{{\text{2}}^{\text{2}}}+\frac{{\text{Gm}}^{\text{2}}}{{\text{4}}^{\text{2}}}\mathrm{......}$

$\text{F}={\text{Gm}}^2\left[\frac{1}{1}+\frac{1}{4}+\frac{1}{4^2}+\mathrm{....}\right]$

$={\text{Gm}}^{\text{2}}\left[1+\frac{1}{4}+\frac{1}{4^2}+\mathrm{....}\right]$         

This is in the form of G.P

Sum of n terms of G.P  ${\text{S}}_{\text{n}}=\frac{\text{a}}{1-\text{r}}$ $\because \text{r}<1$

${\text{S}}_{\text{n}}=\frac{\text{a}}{1-\text{r}}=\frac{1}{1-\text{1/4}}=\text{3/4}$

$\text{F}={\text{Gm}}^{\text{2}}\left[\frac{1}{1-\frac{1}{4}}\right]$

$={\text{Gm}}^{\text{2}}\left[\frac{1}{3/4}\right]$

$={\text{Gm}}^2\left[\frac{4}{3}\right]$

$\text{F}=\frac{4}{3}{\text{Gm}}^{\text{2}}$