JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

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

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the coefficient of

n^3

after expanding the series and neglecting higher powers of

\frac{b}{a}

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