JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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The value of

2\;{C_0} + \frac{{{2^2}}}{2}{C_1} + \frac{{{2^3}}}{3}{C_2} + \frac{{{2^4}}}{4}{C_3} + \ldots + \frac{{{2^{11}}}}{{11}}{C_{10}},

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Correct option: A

\frac{{{3^{11}} - 1}}{{11}}

Option A (

\frac{{{3^{11}} - 1}}{{11}}

) is correct because it represents the sum of the given series, which can be simplified using binomial theorem and integration.

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