Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The coefficient of $x^{50}$ in $(1+x)^{101}\left(1-x+x^2\right)^{100}$ is

Binomial Theorem

Solution:

Here $(1+x)^{101}\left(1-x+x^2\right)^{100}=(1+x)(1+x)^{100}\left(1-x+x^2\right)^{100} $
$=(1+x)\left(1+x^3\right)^{100}$
$=\left(1+x^3\right)^{100}+x\left(1+x^3\right)^{100}$
$\therefore$ coefficient of $x^{50}=0+0=0$