Question

The coefficient of $x^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots \ldots .+x^{500}$ is :

• A. ${ }^{500} C _{301}$
• B. ${ }^{501} C_{200}$
• C. ${ }_3{ }^{500} C_{300}$
• D. ${ }^{501} C_{302}$

Answer 1

Answer: (B)

$ (1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots+x^{500} $
$ =(1+x)^{500} \cdot\left\{\frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}\right\}$
$=(1+x)^{500} \frac{\left((1+x)^{501}-x^{501}\right)}{(1+x)^{501}} \cdot(1+x) $
$=(1+x)^{501}-x^{501}$
Coefficient of $x^{301}$ in $(1+x)^{501}-x^{501}$ is given by
${ }^{501} C_{301}={ }^{501} C_{200}$