If $\mathrm{b}$ is very small as compared to the value of $\mathrm{a},$ so that the cube and other higher powers of $\frac{\mathrm{b}}{\mathrm{a}}$ can be neglected in the identity $\frac{1}{\mathrm{a}-\mathrm{b}}+\frac{1}{2\mathrm{a}-\mathrm{b}}+\frac{1}{\mathrm{a}-3\mathrm{b}}+\dots +\frac{1}{\mathrm{a}-\mathrm{nb}}$

$=\mathrm{\alpha n}+{\mathrm{\beta n}}^2+{\mathrm{\gamma n}}^3.$ Then the value of $\mathrm{\gamma }$ is 

$\begin{array}{rcl} & & \frac{1}{\mathrm{a}} \left[\frac{1}{1-{\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}}+\frac{1}{1-{\displaystyle \frac{2\mathrm{b}}{\mathrm{a}}}}+\frac{1}{1-{\displaystyle \frac{3\mathrm{b}}{\mathrm{a}}}}+\frac{1}{1-{\displaystyle \frac{\mathrm{nb}}{\mathrm{a}}}}\right]\\ & =& \frac{1}{\mathrm{a}} \left({\left(1-\frac{\mathrm{b}}{\mathrm{a}}\right)}^{-1}+{\left(1-2\frac{\mathrm{b}}{\mathrm{a}}\right)}^{-1}+.....\right)\\ & =& \frac{1}{\mathrm{a}} \left[\left(1+\frac{\mathrm{b}}{\mathrm{a}}+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)+\left(1+\frac{2\mathrm{b}}{\mathrm{a}}+\frac{4{\mathrm{b}}^2}{{\mathrm{a}}^2}+....\right)\right]\\ & =& \frac{1}{\mathrm{a}} \left[\mathrm{n}+\frac{\mathrm{b}}{\mathrm{a}}\left(1+2+3+....\mathrm{n}\right)+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2} \left(1^2+2^2+....{\mathrm{n}}^2\right)\right]\\ & =& \frac{1}{\mathrm{a}} \left[\mathrm{n}+\frac{\mathrm{b}}{\mathrm{a}} \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2} \frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6}\right]\\ & =& \frac{1}{\mathrm{a}} \left[\mathrm{n}+\frac{\mathrm{b}}{2\mathrm{a}} \left({\mathrm{n}}^2+\mathrm{n}\right)+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2} \left(\frac{2{\mathrm{n}}^3+3{\mathrm{n}}^2+\mathrm{n}}{6}\right)\right]\\ \mathrm{r}& =& \mathrm{Coeff} {\mathrm{n}}^3=\frac{1}{\mathrm{a}} \left[\frac{{\mathrm{b}}^2}{3{\mathrm{a}}^2}\right]=\frac{{\mathrm{b}}^2}{3{\mathrm{a}}^3}\end{array}$