Mathematics - Binomial Theorem Question with Solution | TestHub

We have,

${\left(1-x\right)}^{101}{\left(x^2+x+1\right)}^{100}$

$={\left(1-x\right)}^{100}{\left(x^2+x+1\right)}^{100}\left(1-x\right)$

$={\left[\left(1-x\right)\left(x^2+x+1\right)\right]}^{100}·\left(1-x\right)$

$={\left(1-x^3\right)}^{100}·\left(1-x\right)$

$=\underset{\mathrm{No} \mathrm{term} \mathrm{of} x^{256}}{\underbrace{{\left(1-{\mathrm{x}}^3\right)}^{100}}}-\underset{\mathrm{we} \mathrm{find} \mathrm{cofficient} \mathrm{of} {\mathrm{x}}^{255}}{\underbrace{\mathrm{x}{\left(1-{\mathrm{x}}^3\right)}^{100}}}$

Now,

${\left(1-x^3\right)}^{100}={}^{100}C_0-{}^{100}C_1{x}^3+{}^{100}C_2{x}^6-...-{}^{100}C_{85}\left(-x^{255}\right)$

$\Rightarrow x{\left(1-x^3\right)}^{100}={}^{100}C_{0}x-{}^{100}C_1{x}^4+{}^{100}C_2{x}^7-...+{}^{100}C_{85}{x}^{256}$

Required coefficient

$={}^{100}C_{85}={}^{100}C_{15}$