If \( b \) is very small as compared to the value of \( a \), so that the cube and other higher powers of \( \dfrac{b}{a} \) can be neglected in the identify
\( \dfrac{1}{a-b} + \dfrac{1}{a-2b} + \dfrac{1}{a-3b} + \ldots + \dfrac{1}{a-nb} = \alpha n + \beta n^2 + \gamma n^3 \), then the value of \( \gamma \) is:
