Question

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

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

Answer 1

Answer: (B)

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