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If $\mathrm{b}$ is very small as compared to the value of $\mathrm{a}$, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b} \ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}$, then the value of $\gamma$ is:
$\frac{b^{2}}{3 a^{3}}$
$\frac{a+b}{3 a^{2}}$
$\frac{a^{2}+b}{3 a^{3}}$
$\frac{a+b^{2}}{3 a^{3}}$
Solution
$(a-b)^{-1}+(a-2 b)^{-1}+\ldots \ldots+(a-n b)^{1}$
$=\frac{1}{a} \sum_{r=1}^{n}\left(1-\frac{r b}{a}\right)^{-1}$ $=\frac{1}{a} \sum_{r=1}^{n}\left\{\left(1+\frac{r b}{a}+\frac{r^{2} b^{2}}{a^{2}}\right)+(\right.$ terms to be neglected $\left.)\right\}$
$=\frac{1}{a}\left[n+\frac{n(n+1)}{2} \cdot \frac{b}{a}+\frac{n(n+1)(2 n+1)}{6} \cdot \frac{b^{2}}{a^{2}}\right]$
$=\frac{1}{a}\left[n^{3}\left(\frac{b^{2}}{3 a^{2}}\right)+\ldots\right]$
So $\gamma=\frac{\mathrm{b}^{2}}{3 \mathrm{a}^{2}}$
