The coefficient of $x^{301}$ in $(1+x)^{500}+x(1+x)^{499}$ $+x^2(1+x)^{498}+\ldots \ldots+x^{500}$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The coefficient of

x^{301}

(1+x)^{500}+x(1+x)^{499}

+x^2(1+x)^{498}+\ldots \ldots+x^{500}

View answer and explanation

Correct option: B

{ }^{501} C_{200}

The coefficient of x³⁰¹ is

{ }^{501} C_{200}

. By summing the geometric series and using the identity

{ }^n C_r + { }^n C_{r+1} = { }^{n+1} C_{r+1}

, the coefficient can be simplified to

{ }^{501} C_{200}

Practice All Chapter MCQs →

\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