JEE Main 2021 (25 Jul, Shift 1) Maths Q70: Binomial Theorem

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

Accepted Answer

Option 4: $\frac{b^2}{3a^3}$

Answer

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

Answer

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

Answer

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

Answer

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

JEE Main 25 July 2021, Shift 1, Maths Q70