JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Value of

\left({ }^{40} C_0\right)\left({ }^{40} C_{15}\right)-\left({ }^{40} C_1\right)\left({ }^{40} C_{16}\right)+\left({ }^{40} C_2\right)\left({ }^{40} C_{12}\right) \cdots

-\left({ }^{40} C_{25}\right)\left({ }^{40} C_{40}\right)

equals

View answer and explanation

Correct option: A

Option A (0) is correct because the given expression represents the coefficient of x²⁵ in (1-x)⁴⁰(1+x)⁴⁰ which simplifies to (1-x²)⁴⁰, and thus the coefficient of x²⁵ is 0.

More Questions from BINOMIAL THEOREM

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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

equals If

a

and

d

are two complex numbers, then the sum to

\left( {n + 1} \right)

terms of the following series

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

is The number of different terms in the expansion of

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

is The sum of the coefficient of the binomial expansion of

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

is equal to 6561 . The constant term in the expansion is The coefficient of

{x^5}

in the expansion of

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

is If

\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

, then

n

equals