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

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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

, then

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Correct option: C

a_{n-3}=a_{n+3}

Option D (

a_{{n-3}}=a_{{n+3}}

) is correct because binomial coefficients are symmetric, meaning

{ }^{2n}C_{{n-3}} = { }^{2n}C_{{n+3}}

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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