JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Analyzing +4 | -1

For

r = 0,\;

…10 let

{A_r},\;{B_r}\;\mathrm{and\;}{C_r}

denotes, respectively, the coefficient of

{x^r}

in the

{\left( {1 + x} \right)^{10}},\;{\left( {1 + x} \right)^{20}},\;

and

{\left( {1 + x} \right)^{30}}

. Then

\sum \limits_{r = 1}^{10} {A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)\mathrm{\;}

is equal to

View answer and explanation

Correct option: D

{C_{10}} - {B_{10}}

Option D (

{C_{10}} - {B_{10}}

) is correct because the summation simplifies to the difference between the coefficients of x¹⁰ in (1+x)³⁰ and (1+x)²⁰.

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