JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Analyzing +4 | -1

The coefficient of

{x^{n - 2}}

in the polynomial

\left( {x - 1} \right)\left( {x - 2} \right) \ldots \left( {x - n} \right)

View answer and explanation

Correct option: B

\frac{1}{{24}}n\left( {{n^2} - 1} \right)\left( {3\;n + 2} \right)

Option B (

\frac{1}{{24}}n\left( {{n^2} - 1} \right)\left( {3\;n + 2} \right)

) is correct because it is the correct coefficient of

{x^{n - 2}}

when expanding the given polynomial.

More Questions from BINOMIAL THEOREM

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