JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

\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

View answer and explanation

Correct option: B

-{ }^{10} C_5

Option B (-

{ }^{10} C_5

) is correct because the given expression is the coefficient of

x^{10}

in the expansion of

(1-x^2)^{10}

, which can be found using the binomial theorem.

More Questions from BINOMIAL THEOREM

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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

{\left( {1 + x} \right)^n} = \sum \limits_{r = 0}^n {C_r}\;{x^r}

\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + \ldots + n\frac{{{C_n}}}{{{C_{n - 1}}}} = \frac{1}{k}n\left( {n + 1} \right),

k