The coefficient of x ^ 301 in (1 + x) ^ 500 + x (1 + x) ^ 499 + x ^...

We have,

${\left(1 + x\right)}^{500} + x{\left(1 + x\right)}^{499} + {x}^{2}{\left(1 + x\right)}^{498} + \ldots . . + {x}^{500}$

This is a geometric progression.

$= {\left(1 + x\right)}^{500} \cdot \left\{\dfrac{1 - {\left(\dfrac{x}{1 + x}\right)}^{501}}{1 - \dfrac{x}{1 + x}}\right\}$

$= {\left(1 + x\right)}^{500}\dfrac{\left({\left(1 + x\right)}^{501} - {x}^{501}\right)}{{\left(1 + x\right)}^{501}} \cdot \left(1 + x\right)$

$= ( 1 + x{)}^{501} - {x}^{501}$

Coefficient of ${x}^{301}$ in $( 1 + x{)}^{501} - {x}^{501}$ is given by

$C301 501 = C200 501$