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 $\frac{1}{a-b} + \frac{1}{a-2b} + \frac{1}{a-3b} + ....... + \frac{1}{a-nb} = \alpha n + \beta n^2 + \gamma n^3$, then the value of $\gamma$ is :
Show Hint
To find the coefficient of the highest power of \(n\) in a summation of polynomial-like terms, you only need to look at the leading term of the last summation component.
Step 1: Understanding the Concept:
Each term in the series can be expanded using the binomial theorem for negative powers because \(b/a\) is small.
We sum these individual expansions up to \(n\) terms and collect the terms involving \(n\), \(n^2\), and \(n^3\). Step 2: Key Formula or Approach:
1. \((1 - x)^{-1} = 1 + x + x^2 + .......\) for small \(x\).
2. \(\sum_{k=1}^n k = \frac{n(n+1)}{2} \approx \frac{n^2}{2}\).
3. \(\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3}\). Step 3: Detailed Explanation:
The general term is \(T_k = \frac{1}{a - kb} = \frac{1}{a(1 - \frac{kb}{a})} = \frac{1}{a} (1 - \frac{kb}{a})^{-1}\).
Expanding: \(T_k \approx \frac{1}{a} [1 + \frac{kb}{a} + (\frac{kb}{a})^2]\). (Higher powers are neglected).
Sum \(S = \sum_{k=1}^n T_k = \frac{1}{a} \sum_{k=1}^n [1 + \frac{kb}{a} + \frac{k^2b^2}{a^2}]\).
\(S = \frac{1}{a} [\sum 1 + \frac{b}{a} \sum k + \frac{b^2}{a^2} \sum k^2]\).
Using sum formulas:
\(S = \frac{1}{a} [n + \frac{b}{a} \frac{n(n+1)}{2} + \frac{b^2}{a^2} \frac{n(n+1)(2n+1)}{6}]\).
The term with \(n^3\) comes from the expansion of \(\frac{b^2}{a^3} \frac{2n^3 + .......}{6}\).
The coefficient of \(n^3\) is \(\gamma = \frac{b^2}{a^3} \cdot \frac{2}{6} = \frac{b^2}{3a^3}\). Step 4: Final Answer:
The value of \(\gamma\) is \(\frac{b^2}{3a^3}\).
