JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The value of

\left(\begin{array}{l}30 \\ 0\end{array}\right)\left(\begin{array}{l}30 \\ 10\end{array}\right)-\left(\begin{array}{l}30 \\ 1\end{array}\right)\left(\begin{array}{l}30 \\ 11\end{array}\right)

+\left(\begin{array}{l}30 \\ 2\end{array}\right)\left(\begin{array}{l}30 \\ 12\end{array}\right)+\ldots+\left(\begin{array}{l}30 \\ 20\end{array}\right)\left(\begin{array}{l}30 \\ 30\end{array}\right)

is equal to

View answer and explanation

Correct option: B

{ }^{30} C_{10}

Option B (

{ }^{30} C_{10}

) is correct because the given expression is the coefficient of

x^{20}

in the product of

(1-x)^{30}

and

(x+1)^{30}

, which simplifies to

(1-x^2)^{30}

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