The coefficient of x 101 in the expression 5 + x 500 + x 5 + x 499 + x 2 5 + x 498 + … … + x 500…

Given, 

${\left(5+x\right)}^{500}+x{\left(5+x\right)}^{499}+x^2{\left(5+x\right)}^{498}+\dots \dots +x^{500}$

Using sum of the G.P in the given expression we get,

$\Rightarrow \frac{{\left(5+x\right)}^{500}\left[{\left(\frac{x}{5+x}\right)}^{501}-1\right]}{\left(\frac{x}{5+x}\right)-1}=\frac{x^{501}-{\left(5+x\right)}^{501}}{-5}$

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

Now coefficient of $x^{101}$ in $\frac{1}{5}{\left(5+x\right)}^{501}$ will be $\frac{1}{5}\left({}^{501}C_{101}\right){\left(5\right)}^{501-101}$$=5^{399}\left({}^{501}C_{101}\right)$