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

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The coefficient of

{x^{256}}

in the expansion of

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

View answer and explanation

Correct option: B

^{100}{C_{15}}

The correct answer is B (

^{100}{C_{15}}

) because it is the coefficient of

{x^{256}}

obtained after simplifying the expression using the identity

{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})

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\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n