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Step-by-step Solution
Step 1: Express the given sum
Consider the sum
$S = \frac{1}{a - b} + \frac{1}{a - 2b} + \frac{1}{a - 3b} + \dots + \frac{1}{a - nb}.$
Step 2: Factor out $1/a$ from each term
Notice that each denominator has the form $a - rb$ (where $r$ goes from 1 to $n$).
We factor $a$ out of the denominator:
$$
\frac{1}{a - rb}
= \frac{1}{a\Bigl(1 - \frac{rb}{a}\Bigr)}
= \frac{1}{a} \cdot \frac{1}{\bigl(1 - \frac{rb}{a}\bigr)}.
$$
Hence,
$$
S
= \sum_{r=1}^n \frac{1}{a - rb}
= \frac{1}{a} \sum_{r=1}^n \frac{1}{\left(1 - \frac{rb}{a}\right)}.
$$
Step 3: Use binomial expansion for small $b/a$
Because $b$ is very small compared to $a$, $\bigl(\tfrac{b}{a}\bigr)^2$ and higher powers are progressively smaller.
We use the expansion for
$
\frac{1}{1 - x}
\approx 1 + x + x^2 + x^3 + \dots
$
and keep terms up to $\bigl(\tfrac{b}{a}\bigr)^2$ (since higher powers are neglected by the problem statement).
Therefore, for each term:
$$
\frac{1}{1 - \frac{rb}{a}}
\approx 1 + \frac{rb}{a} + \Bigl(\frac{rb}{a}\Bigr)^2.
$$
Step 4: Sum the series term-by-term
Substituting back, we get
$$
\sum_{r=1}^n \frac{1}{1 - \frac{rb}{a}}
\approx \sum_{r=1}^n
\Bigl[
1 + \frac{rb}{a} + \Bigl(\frac{rb}{a}\Bigr)^2
\Bigr].
$$
Break this into separate sums:
$$
\sum_{r=1}^n 1 = n,
$$
$$
\sum_{r=1}^n \frac{rb}{a}
= \frac{b}{a} \sum_{r=1}^n r
= \frac{b}{a} \cdot \frac{n(n+1)}{2},
$$
$$
\sum_{r=1}^n \Bigl(\frac{rb}{a}\Bigr)^2
= \frac{b^2}{a^2} \sum_{r=1}^n r^2
= \frac{b^2}{a^2} \cdot \frac{n(n+1)(2n+1)}{6}.
$$
Hence,
$$
\sum_{r=1}^n \frac{1}{1 - \frac{rb}{a}}
\approx
n
+ \frac{b}{a}\,\frac{n(n+1)}{2}
+ \frac{b^2}{a^2}\,\frac{n(n+1)(2n+1)}{6}.
$$
Step 5: Reintroduce the factor $1/a$
Recall from Step 2 that
$
S = \frac{1}{a} \sum_{r=1}^n \frac{1}{\left(1 - \frac{rb}{a}\right)}.
$
Therefore,
$$
S \approx \frac{1}{a} \biggl[
n
+ \frac{b}{a}\,\frac{n(n+1)}{2}
+ \frac{b^2}{a^2}\,\frac{n(n+1)(2n+1)}{6}
\biggr].
$$
When this sum is represented in the form
$
\alpha\,n + \beta\,n^2 + \gamma\,n^3,
$
the coefficient of $n^3$ emerges from the term with $\frac{b^2}{a^2} \cdot \frac{n(n+1)(2n+1)}{6} \cdot \frac{1}{a}$, which simplifies to
$$
\frac{b^2}{a^3} \times \frac{n(n+1)(2n+1)}{6}.
$$
The leading behavior in $n^3$ has the coefficient
$$
\gamma = \frac{b^2}{3\,a^3}.
$$
Thus, the required value of $\gamma$ is
$$
\gamma = \frac{b^2}{3a^3}.
$$
