The coefficient of

Question:

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

  1. ${ }^{100} \mathrm{C}_{16}$

  2. ${ }^{100} \mathrm{C}_{15}$

  3. $-{ }^{100} \mathrm{C}_{16}$

  4. $-{ }^{100} \mathrm{C}_{15}$


Correct Option: , 2

Solution:

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

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

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

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

$=\underbrace{\left(1-x^{3}\right)^{100}}_{\text {Notermof } x^{256}}-\underbrace{x}_{\text {We findcofficientof } x^{255}}$

Required coefficient $(-1) \times\left(-{ }^{100} \mathrm{C}_{85}\right)$

$={ }^{100} \mathrm{C}_{85}={ }^{100} \mathrm{C}_{15}$

Leave a comment