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 $$ = {1 \over 5}\,.\,{}^{501}{C_{101}}\,.\,{5^{400}}$$

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