Question

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

• A. $\frac{b^2}{3a^3}$
• B. $\frac{b^2}{2a^3}$
• C. $\frac{b}{a^3}$
• D. $\frac{b^3}{a^3}$

Answer 1

Answer: (A)

$\frac{b^2}{3a^3}$

Answer 2

The general term $\frac{1}{a-kb}$ is expanded using binomial approximation: $\frac{1}{a-kb} = \frac{1}{a(1 - \frac{kb}{a})} = \frac{1}{a} \left(1 - \frac{kb}{a}\right)^{-1}$ Given that $b$ is very small compared to $a$, we neglect higher powers of $\frac{b}{a}$. Using the binomial expansion $(1-x)^{-1} \approx 1 + x + x^2$ for small $x$: $\left(1 - \frac{kb}{a}\right)^{-1} \approx 1 + \frac{kb}{a} + \left(\frac{kb}{a}\right)^2 = 1 + \frac{kb}{a} + \frac{k^2b^2}{a^2}$ So, the general term is approximated as: $\frac{1}{a-kb} \approx \frac{1}{a} \left(1 + \frac{kb}{a} + \frac{k^2b^2}{a^2}\right) = \frac{1}{a} + \frac{kb}{a^2} + \frac{k^2b^2}{a^3}$ Now, we sum this approximation from $k=1$ to $n$: $\sum_{k=1}^{n} \frac{1}{a-kb} \approx \sum_{k=1}^{n} \left(\frac{1}{a} + \frac{kb}{a^2} + \frac{k^2b^2}{a^3}\right)$ $= \frac{1}{a} \sum_{k=1}^{n} 1 + \frac{b}{a^2} \sum_{k=1}^{n} k + \frac{b^2}{a^3} \sum_{k=1}^{n} k^2$ Using the standard summation formulas: $\sum_{k=1}^{n} 1 = n$ $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ Substituting these into the sum: $\text{Sum} \approx \frac{1}{a}(n) + \frac{b}{a^2}\left(\frac{n(n+1)}{2}\right) + \frac{b^2}{a^3}\left(\frac{n(n+1)(2n+1)}{6}\right)$ We are given the identity $\alpha n + \beta n^{2} + \gamma n^{3}$. We need to find the coefficient of $n^3$. Let's expand the sum of squares term: $\frac{n(n+1)(2n+1)}{6} = \frac{n(2n^2 + 3n + 1)}{6} = \frac{2n^3 + 3n^2 + n}{6} = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n$ So the third term of the sum is: $\frac{b^2}{a^3}\left(\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n\right) = \frac{b^2}{3a^3}n^3 + \frac{b^2}{2a^3}n^2 + \frac{b^2}{6a^3}n$ The full sum, when expanded and collected by powers of $n$, will have the form: $\text{Sum} \approx \left(\frac{b^2}{3a^3}\right)n^3 + \left(\frac{b}{2a^2} + \frac{b^2}{2a^3}\right)n^2 + \left(\frac{1}{a} + \frac{b}{2a^2} + \frac{b^2}{6a^3}\right)n$ Comparing this to $\alpha n + \beta n^{2} + \gamma n^{3}$, the coefficient of $n^3$ is $\gamma$. Therefore, $\gamma = \frac{b^2}{3a^3}$.