Key Concept: Binomial Approximation for Small Values

This problem utilizes the binomial theorem for negative integer exponents, specifically the approximation for $(1-x)^{-1}$ when $x$ is very small. The general binomial expansion is: $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \ldots$ For $(1-x)^{-1}$, we set $y = -x$ and $n = -1$: $(1-x)^{-1} = 1 + (-1)(-x) + \frac{(-1)(-2)}{2!}(-x)^2 + \frac{(-1)(-2)(-3)}{3!}(-x)^3 + \ldots$ $(1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots$ The problem states that $b$ is very small compared to $a$, allowing us to neglect the cube and other higher powers of $b/a$. This means that if $x = k \cdot (b/a)$, then $(b/a)^3$ and higher order terms involving $b/a$ are considered negligible. Therefore, we will approximate $(1-x)^{-1}$ as: $(1-x)^{-1} \approx 1 + x + x^2$

Rewriting Each Term of the Sum

The given series is: $S = {1 \over {a - b}} + {1 \over {a - 2b}} + {1 \over {a - 3b}} + \ldots + {1 \over {a - nb}}$ We can write the general $r$-th term of this sum as $T_r = \frac{1}{a - rb}$, where $r$ ranges from $1$ to $n$. To apply the binomial approximation, we first factor out $a$ from the denominator of each term: $T_r = {1 \over {a - rb}} = {1 \over {a\left( {1 - {{rb} \over a}} \right)}} = {1 \over a}{\left( {1 - {{rb} \over a}} \right)^{ - 1}}$ Here, $x = \frac{rb}{a}$. Since $b/a$ is very small, $rb/a$ will also be very small for reasonable values of $r$. According to the problem statement, we can neglect terms with $(b/a)^3$ and higher powers. This implies we need to expand up to the $(rb/a)^2$ term. Applying the binomial approximation $(1-x)^{-1} \approx 1 + x + x^2$ with $x = \frac{rb}{a}$: ${\left( {1 - {{rb} \over a}} \right)^{ - 1}} \approx 1 + {{rb} \over a} + {\left( {{{rb} \over a}} \right)^2} = 1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}$ So, each term $T_r$ can be approximated as: $T_r \approx {1 \over a}\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right)$

Summation of the Series

Now, we sum these approximated terms from $r=1$ to $n$: $S = \sum\limits_{r = 1}^n {T_r} = \sum\limits_{r = 1}^n {{1 \over a}\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right)}$ We can factor out the constant $1/a$ from the summation: $S = {1 \over a}\sum\limits_{r = 1}^n {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right)}$ Distribute the summation operator: $S = {1 \over a}\left[ {\sum\limits_{r = 1}^n 1 + \sum\limits_{r = 1}^n {{{rb} \over a}} + \sum\limits_{r = 1}^n {{{{r^2}{b^2}} \over {{a^2}}}}} \right]$ Factor out constants from each individual summation: $S = {1 \over a}\left[ {\sum\limits_{r = 1}^n 1 + {b \over a}\sum\limits_{r = 1}^n r + {{{b^2}} \over {{a^2}}}\sum\limits_{r = 1}^n {{r^2}}} \right]$ Now, we use the standard summation formulas:

1. $\sum\limits_{r = 1}^n 1 = n$
2. $\sum\limits_{r = 1}^n r = {{n(n + 1)} \over 2}$
3. $\sum\limits_{r = 1}^n {{r^2}} = {{n(n + 1)(2n + 1)} \over 6}$ Substitute these into the expression for $S$: $S = {1 \over a}\left[ {n + {b \over a}\frac{n(n+1)}{2} + {{{b^2}} \over {{a^2}}}\frac{n(n+1)(2n+1)}{6}} \right]$

Extracting the Coefficient of $n^3$

The problem asks for the value of $\gamma$, which is the coefficient of $n^3$ in the expanded form of $S = \alpha n + \beta {n^2} + \gamma {n^3}$. We need to expand the expression we derived for $S$ and identify the $n^3$ term. Let's look at each term inside the large square bracket:

1. The first term is $n$. This contributes to the $n^1$ term, not $n^3$.
2. The second term is ${b \over a}\frac{n(n+1)}{2}$. Expanding this gives: ${b \over {2a}}(n^2 + n) = {{bn^2} \over {2a}} + {{bn} \over {2a}}$ This term contributes to $n^2$ and $n^1$, not $n^3$.
3. The third term is ${{{b^2}} \over {{a^2}}}\frac{n(n+1)(2n+1)}{6}$. To find the $n^3$ component, we expand the polynomial in $n$: $n(n+1)(2n+1) = (n^2+n)(2n+1) = 2n^3 + n^2 + 2n^2 + n = 2n^3 + 3n^2 + n$ So, the third term becomes: ${{{b^2}} \over {{a^2}}}\frac{2n^3 + 3n^2 + n}{6} = {{{b^2}} \over {{a^2}}}\left( {{{2n^3}} \over 6} + {{{3n^2}} \over 6} + {{n} \over 6} \right) = {{{b^2}} \over {{a^2}}}\left( {{{n^3}} \over 3} + {{{n^2}} \over 2} + {{n} \over 6} \right)$ From this term, the component containing $n^3$ is ${{{b^2}} \over {{a^2}}} \cdot {{{n^3}} \over 3} = {{{b^2}n^3} \over {3{a^2}}}$.

Combining these, only the third term contributes to $n^3$ inside the bracket. So, the coefficient of $n^3$ inside the bracket is $\frac{b^2}{3a^2}$. Finally, we must remember the leading factor of $\frac{1}{a}$ that was factored out initially: $S = {1 \over a}\left[ \ldots + \left( {{{b^2}} \over {3{a^2}}} \right)n^3 + \ldots \right]$ Therefore, the coefficient of $n^3$ in the full expansion of $S$, which is $\gamma$, is: $\gamma = {1 \over a} \cdot {{{b^2}} \over {3{a^2}}} = {{{b^2}} \over {3{a^3}}}$

Tips for Solving Similar Problems

• Understand the Approximation Condition: Pay close attention to which terms can be neglected. Here, "cube and higher powers of $b/a$" meant expanding only up to $(b/a)^2$.
• Systematic Expansion: Always write out the binomial expansion step-by-step. Avoid mental shortcuts, especially when dealing with multiple small terms like $rb/a$.
• Correct Summation Formulas: Memorize or have readily available the sum formulas for $\sum 1$, $\sum r$, and $\sum r^2$. Errors here are common.
• Algebraic Precision: Be careful when expanding polynomial terms like $n(n+1)(2n+1)$ and collecting coefficients.
• Don't Forget External Factors: Ensure you multiply by any initial factors (like $1/a$ in this case) at the very end to get the final coefficient.

Summary and Key Takeaway

This problem effectively combines the application of binomial approximations for terms with small ratios and the summation of series using standard formulas. By systematically expanding each term, collecting them, and then identifying the coefficient of the desired power of $n$, we find that $\gamma = {{{b^2}} \over {3{a^3}}}$.