The coefficient of $x^{256}$ in the expansion of $(1-x)^{101}$ $\left(x^{2}+x+1 ight)^{100}$ is\

Question

The coefficient of $x^{256}$ in the expansion of $(1-x)^{101}$ $\left(x^{2}+x+1ight)^{100}$ is\

Professor PC Thomas Classes · Answer

(b) : We have, $(1-x)^{101} \cdot\left(x^{2}+x+1ight)^{100}$ $=(1-x)^{100} \cdot(1-x)\left(x^{2}+x+1ight)^{100}$\ $=\left((1-x)\left(x^{2}+x+1ight)ight)^{100} \cdot(1-x)$\ $=\left(1-x^{3}ight)^{100} \cdot(1-x)$\ $=\left(1-x^{3}ight)^{100}-x\left(1-x^{3}ight)^{100}$\ $	herefore \quad$ Required coefficient\ $=(-1) 	imes\left(-{ }^{100} C_{85}ight)={ }^{100} C_{85}={ }^{100} C_{15}$\