QJEEM21Ma3aed: JEE Main 2021 Mathematics - Class 11 JEEM21Ma3aed Solution: If b is very small as compared to the value of a, so that the…

Question

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If b is very small as compared to the value of a, so that the cube and other higher powers of

{b \over a}

can be neglected in the identity

{1 \over {a - b}} + {1 \over {a - 2b}} + {1 \over {a - 3b}} + ..... + {1 \over {a - nb}} = \alpha n + \beta {n^2} + \gamma {n^3}

, then the value of

\gamma

is :

Accepted Answer

{(a - b)^{ - 1}} + {(a - 2b)^{ - 1}} + .... + {(a - nb)^{ - 1}}

= {1 \over a}\sum\limits_{r = 1}^n {{{\left( {1 - {{rb} \over a}} \right)}^{ - 1}}}

= {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \right\}}

= {1 \over a}\left[ {n + {{n(n + 1)} \over 2}.{b \over a} + {{n(n + 1)(2n + 1)} \over 6}.{{{b^2}} \over {{a^2}}}} \right]

= {1 \over a}\left[ {{n^3}\left( {{{{b^2}} \over {3{a^2}}}} \right) + .....} \right]

So,

\gamma = {{{b^2}} \over {3{a^3}}}

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