JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 6)
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 :
$${{{a^2} + b} \over {3{a^3}}}$$
$${{a + b} \over {3{a^2}}}$$
$${{{b^2}} \over {3{a^3}}}$$
$${{a + {b^2}} \over {3{a^3}}}$$
Explanation
$${(a - b)^{ - 1}} + {(a - 2b)^{ - 1}} + .... + {(a - nb)^{ - 1}}$$
$$ = {1 \over a}\sum\limits_{r = 1}^n {{{\left( {1 - {{rb} \over a}} \right)}^{ - 1}}} $$
$$ = {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \right\}} $$
$$ = {1 \over a}\left[ {n + {{n(n + 1)} \over 2}.{b \over a} + {{n(n + 1)(2n + 1)} \over 6}.{{{b^2}} \over {{a^2}}}} \right]$$
$$ = {1 \over a}\left[ {{n^3}\left( {{{{b^2}} \over {3{a^2}}}} \right) + .....} \right]$$
So, $$\gamma = {{{b^2}} \over {3{a^3}}}$$
$$ = {1 \over a}\sum\limits_{r = 1}^n {{{\left( {1 - {{rb} \over a}} \right)}^{ - 1}}} $$
$$ = {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \right\}} $$
$$ = {1 \over a}\left[ {n + {{n(n + 1)} \over 2}.{b \over a} + {{n(n + 1)(2n + 1)} \over 6}.{{{b^2}} \over {{a^2}}}} \right]$$
$$ = {1 \over a}\left[ {{n^3}\left( {{{{b^2}} \over {3{a^2}}}} \right) + .....} \right]$$
So, $$\gamma = {{{b^2}} \over {3{a^3}}}$$
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