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

(D) ${ }^{100} \mathrm{C}_{15}$

(A) $-{ }^{100} \mathrm{C}_{15}$

(B) ${ }^{100} \mathrm{C}_{16}$

(C) $-{ }^{100} \mathrm{C}_{16}$

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Binomial Theorem

The coefficient of

x^{256}

in the expansion of

(1 - x)^{101} (x^{2} + x + 1)^{100}

is :

Answer: (D)

Since

(x - 1) (x^{2} + x + 1) = x^{3} - 1

(1 - x)^{101} (x^{2} + x + 1)^{100} = - (x - 1)^{101} \cdot \frac{( x ^{3} - 1 ) ^{100}}{( x - 1 ) ^{100}} = - (x - 1) (x^{3} - 1)^{100} = (1 - x^{3})^{100} - x (1 - x^{3})^{100}

using that an even power makes

(x^{3} - 1)^{100} = (1 - x^{3})^{100}

. In

(1 - x^{3})^{100} = \sum_{k}^{100} C_{k} (- 1)^{k} x^{3 k}

only exponents divisible by

3

appear, and

256

is not one of them, so the first of the two pieces contributes nothing to

x^{256}

. The second piece needs the coefficient of

x^{255} = x^{3 (85)}

, which is

^{100} C_{85} (- 1)^{85} = -^{100} C_{85}

, and the factor

- x

turns it into

+^{100} C_{85}

. The coefficient of

x^{256}

is therefore

^{100} C_{15}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option