QJEEM22Ma785a: JEE Main 2022 Mathematics - Class 11 JEEM22Ma785a Solution: The coefficient of x 101 in the expression , x > 0, is

Given,

${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}}\,\, +$ ...... ${x^{500}}$

This is a G.P. with first term ${(5 + x)^{500}}$

Common ratio $= {{x{{(5 + x)}^{499}}} \over {{{(5 + x)}^{500}}}} = {x \over {5 + x}}$ and 501 terms present.

$\therefore$ Sum $= {{{{(5 + x)}^{500}}\left( {{{\left( {{x \over {5 + x}}} \right)}^{501}} - 1} \right)} \over {{x \over {5 + x}} - 1}}$

$= {{{{(5 + x)}^{500}}\left( {{{{x^{501}} - {{(5 + x)}^{501}}} \over {{{(5 + x)}^{501}}}}} \right)} \over {{{x - 5 - x} \over {5 + x}}}}$

$= {{{{{x^{501}} - {{(5 + x)}^{501}}} \over {5 + x}}} \over {{{ - 5} \over {5 + x}}}}$

$= {1 \over 5}\left( {{{\left( {5 + x} \right)}^{501}} - {x^{501}}} \right)$

Coefficient of x¹⁰¹ in ${(5 + x)^{501}}$ is $= {}^{501}{C_{101}}\,.\,{5^{400}}$

$\therefore$ In ${1 \over 5}\left( {{{(5 + x)}^{500}} - {x^{501}}} \right)$ coefficient of x¹⁰¹ is $= {1 \over 5}\,.\,{}^{501}{C_{101}}\,.\,{5^{400}}$

$= {}^{501}{C_{101}}\,.\,{5^{399}}$