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

We have to evaluate the sum

$$S=\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\ldots+\frac{1}{a-nb}$$

under the assumption that the quantity $$b$$ is very small compared with $$a$$. In other words $$\left|\dfrac{b}{a}\right|\ll1$$, so cubes and higher powers of $$\dfrac{b}{a}$$ may be ignored. The question tells us that after such an approximation the sum can be written in the cubic form

$$S=\alpha n+\beta n^{2}+\gamma n^{3},$$

and we have to find the coefficient $$\gamma$$.

To begin, look at a general term of the sum:

$$\frac{1}{a-kb},\qquad k=1,2,3,\ldots ,n.$$

We first factor out $$a$$ from the denominator:

$$\frac{1}{a-kb}=\frac{1}{a}\cdot\frac{1}{1-\dfrac{kb}{a}}.$$

Now we use the binomial (geometric-series) expansion for the reciprocal, valid for $$|x|\lt1$$:

$$\frac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots.$$

Take $$x=\dfrac{kb}{a}$$. Because $$\dfrac{b}{a}$$ is small we retain only the terms up to $$x^{2}$$, discarding $$x^{3}$$ and higher:

$$\frac{1}{1-\dfrac{kb}{a}}\approx1+\frac{kb}{a}+\left(\frac{kb}{a}\right)^{2}.$$

Substituting this back, the single term becomes

$$\frac{1}{a-kb}\approx\frac{1}{a}\Bigl[\,1+\frac{kb}{a}+\frac{k^{2}b^{2}}{a^{2}}\,\Bigr].$$

Now we sum this expression for $$k=1$$ to $$k=n$$:

$$S\approx\frac{1}{a}\sum_{k=1}^{n}\biggl[\,1+\frac{kb}{a}+\frac{k^{2}b^{2}}{a^{2}}\,\biggr].$$

We separate the three sums:

$$S\approx\frac{1}{a}\Biggl[\;\underbrace{\sum_{k=1}^{n}1}_{\text{sum of 1's}}+\frac{b}{a}\underbrace{\sum_{k=1}^{n}k}_{\text{sum of first }n\text{ integers}}+\frac{b^{2}}{a^{2}}\underbrace{\sum_{k=1}^{n}k^{2}}_{\text{sum of squares}}\Biggr].$$

We now insert the standard summation formulas:

1. Sum of ones: $$\displaystyle\sum_{k=1}^{n}1=n.$$

2. Sum of the first $$n$$ natural numbers: $$\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$$

3. Sum of the squares of the first $$n$$ natural numbers: $$\displaystyle\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}.$$

Substituting these results into the expression for $$S$$ we get

$$S\approx\frac{1}{a}\Biggl[n+\frac{b}{a}\cdot\frac{n(n+1)}{2}+\frac{b^{2}}{a^{2}}\cdot\frac{n(n+1)(2n+1)}{6}\Biggr].$$

Next we expand each bracket so that every term is expressed as a power of $$n$$.

First note that

$$\frac{n(n+1)}{2}=\frac{n^{2}+n}{2}.$$

Also,

$$\frac{n(n+1)(2n+1)}{6}=\frac{2n^{3}+3n^{2}+n}{6}.$$

Using these, the sum becomes

$$S\approx\frac{1}{a}\Biggl[n+\frac{b}{a}\cdot\frac{n^{2}+n}{2}+\frac{b^{2}}{a^{2}}\cdot\frac{2n^{3}+3n^{2}+n}{6}\Biggr].$$

Now we distribute the outer factor $$\dfrac{1}{a}$$ and arrange the result as a polynomial in $$n$$:

$$S\approx\frac{n}{a}\;+\;\frac{b}{a^{2}}\cdot\frac{n^{2}+n}{2}\;+\;\frac{b^{2}}{a^{3}}\cdot\frac{2n^{3}+3n^{2}+n}{6}.$$

We separate the individual powers of $$n$$ explicitly.

For the $$n^{3}$$ term:

$$\frac{b^{2}}{a^{3}}\cdot\frac{2n^{3}}{6}=\frac{b^{2}}{a^{3}}\cdot\frac{n^{3}}{3}=\frac{b^{2}n^{3}}{3a^{3}}.$$

For the $$n^{2}$$ term:

$$\frac{b}{a^{2}}\cdot\frac{n^{2}}{2}+\frac{b^{2}}{a^{3}}\cdot\frac{3n^{2}}{6} =\frac{b n^{2}}{2a^{2}}+\frac{b^{2}n^{2}}{2a^{3}}.$$

For the $$n$$ term:

$$\frac{n}{a}+\frac{b}{a^{2}}\cdot\frac{n}{2}+\frac{b^{2}}{a^{3}}\cdot\frac{n}{6} =\frac{n}{a}+\frac{b n}{2a^{2}}+\frac{b^{2} n}{6a^{3}}.$$

Collecting, the polynomial form of $$S$$ up to the cubic term is therefore

$$S=\Bigl[\frac{1}{a}+\frac{b}{2a^{2}}+\frac{b^{2}}{6a^{3}}\Bigr]n \;+\;\Bigl[\frac{b}{2a^{2}}+\frac{b^{2}}{2a^{3}}\Bigr]n^{2} \;+\;\Bigl[\frac{b^{2}}{3a^{3}}\Bigr]n^{3}.$$

Comparing with the required expression $$S=\alpha n+\beta n^{2}+\gamma n^{3},$$ we directly read

$$\gamma=\frac{b^{2}}{3a^{3}}.$$

This matches Option C of the given choices.

Hence, the correct answer is Option C.