JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

(1+x)^{2 n}=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}

, then

\left(1+\frac{a_1}{a_0}\right)\left(1+\frac{a_2}{a_1}\right)\left(1+\frac{a_3}{a_2}\right) \ldots\left(1+\frac{a_{2 n}}{a_{2 n-1}}\right)

is equal to

View answer and explanation

Correct option: B

\frac{(2 n+1)^{2 n}}{(2 n) !}

Option B (

\frac{{(2 n+1)^{2 n}}}{{(2 n) !}}

) is correct because it is obtained by simplifying the product of terms involving binomial coefficients.

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

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