JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

Sum of the first 20 terms of the series

\frac{1}{(2)(4)}+\frac{(1)(3)}{(2)(4)(6)}+\frac{(1)(3)(5)}{(2)(4)(6)(8)}+\ldots

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Correct option: C

\frac{1}{2}-\frac{1}{2^{42}}\left({ }^{42} C_{21}\right)

Option C,

\frac{{1}}{{2}}-\frac{{1}}{{2^{42}}}\left({ }^{{42}} C_{{21}}\right)

, is correct because the series can be expressed as a telescoping sum, and the sum of the first 20 terms is equal to

b_1 - b_{21}

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