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 $\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b}+\ldots \ldots+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}$ then the value of $y$ is :

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

Answer 1

Answer: (C)

$(a-b)^{-1}+(a-2 b)^{-1}+\ldots \ldots+(a-n b)^{-1}$
$=\frac{1}{a} \displaystyle\sum_{r=1}^{n}\left(1-\frac{r b}{a}\right)^{-1}$
$=\frac{1}{a} \displaystyle\sum_{r=1}^{n}\left\{\left(1+\frac{r b}{a}+\frac{r^{2} b^{2}}{a^{2}}\right)+(\right.$ terms to be neglected $\left.)\right\}$
$=\frac{1}{a}\left[n+\frac{n(n+1)}{2} \cdot \frac{b}{a}+\frac{n(n+1)(2 n+1)}{6} \cdot \frac{b^{2}}{a^{2}}\right]$
$=\frac{1}{a}\left[n^{3}\left(\frac{b^{2}}{3 a^{2}}\right)+\ldots \ldots\right]$
So $\gamma=\frac{b^{2}}{3 a^{3}}$