The coefficient of ${x^m}$ in ${\left( {1 + x} \right)^p} + {\left( {1 + x} \right)^{p + 1}} + \ldots + {\left( {1 + x} \right)^n},p \le m \le n$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Application +4 | -1

The coefficient of

{x^m}

{\left( {1 + x} \right)^p} + {\left( {1 + x} \right)^{p + 1}} + \ldots + {\left( {1 + x} \right)^n},p \le m \le n

View answer and explanation

Correct option: A

{\;^{n + 1}}{C_{m + 1}}

The correct answer is

{\;^{n + 1}}{C_{m + 1}}

because the coefficient of

{x^m}

can be found by manipulating the sum into a single term and then finding the coefficient of

{x^{m+1}}

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

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