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  • If is very small as compared to the value of a, so that the cube and other higher powers of <img alt="\frac{b}{a}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7Bb%7D%7Ba%

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-2 b}+\frac{1}{a-3 b}+\ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}, then the value of \gamma is
Option: 1 \frac{a^{2}+b}{3 a^{3}}
Option: 2 \frac{a^{2}+b}{3 a^{3}}
Option: 3 \frac{a^{2}+b}{3 a^{3}}
Option: 4 \frac{a^{2}+b}{3 a^{3}}
Option: 5 \frac{a+b}{3 a^{2}}
Option: 6 \frac{a+b}{3 a^{2}}
Option: 7 \frac{a+b}{3 a^{2}}
Option: 8 \frac{a+b}{3 a^{2}}
Option: 9 \frac{b^{2}}{3 a^{3}}
Option: 10 \frac{b^{2}}{3 a^{3}}
Option: 11 \frac{b^{2}}{3 a^{3}}
Option: 12 \frac{b^{2}}{3 a^{3}}
Option: 13 \frac{a+b^{2}}{3 a^{3}}
Option: 14 \frac{a+b^{2}}{3 a^{3}}
Option: 15 \frac{a+b^{2}}{3 a^{3}}
Option: 16 \frac{a+b^{2}}{3 a^{3}}

Answers (1)

best_answer

\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+---+\frac{1}{a-nb}
= \left ( a-b \right )^{-1}+\left ( a-2b \right )^{-1}+\left ( a-3b \right )^{-1}+---\left ( a-nb \right )^{-1}
=a^{-1}\left [ \left ( 1-\frac{b}{a} \right )^{-1} +\left ( 1-\frac{2b}{a} \right )^{-1}+----+\left ( 1-\frac{nb}{a} \right )^{-1}\right ]
=a^{-1}\left [ \left ( 1+\frac{b}{a}+\frac{b^{2}}{a^{2}} \right ) +\left ( 1+\frac{2b}{a}+\frac{4b^{2}}{a^{2}} \right )+--+\left ( 1+\frac{nb}{a} +\frac{n^{2}b^{2}}{a^{2}}\right )\right ]
=\frac{1}{a}\left [ n+\frac{b}{a}\frac{n\left ( n+1 \right )}{2}+\frac{b^{2}}{a}\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6} \right ]

so, \gamma = coefficent of n^{3} in this expression
= \frac{1}{a}\cdot \frac{b^{2}}{3a^{2}}= \frac{b^{2}}{3a^{3}}

Posted by

Kuldeep Maurya

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