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

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

Answer 1

Answer: (C)

$(a-b{)}^{-1}+(a-2b{)}^{-1}+\dots \dots +(a-nb{)}^{-1}$
$=\frac{1}{a}\sum _{r=1}^n{\left(1-\frac{rb}{a}\right)}^{-1}$
$=\frac{1}{a}\sum _{r=1}^n\{\left(1+\frac{rb}{a}+\frac{r^2{b}^2}{a^2}\right)+($ terms to be neglected $)\}$
$=\frac{1}{a}\left[n+\frac{n(n+1)}{2}\cdot \frac{b}{a}+\frac{n(n+1)(2n+1)}{6}\cdot \frac{b^2}{a^2}\right]$
$=\frac{1}{a}\left[n^3\left(\frac{b^2}{3a^2}\right)+\dots \dots \right]$
So $\gamma =\frac{b^2}{3a^3}$