JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The sum of the series

\sum \limits_{r = 0}^n {\left( { - 1} \right)^r}{\;^n}{C_r}\left( {\frac{1}{{{2^r}}} + \frac{{{3^r}}}{{{2^{2r}}}} + \frac{{{7^r}}}{{{2^{3r}}}} + \frac{{{{15}^r}}}{{{2^{4r}}}} + ... + m\mathrm{\;terms}} \right)

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Correct option: A

\frac{{{2^{mn}} - 1}}{{{2^{mn}}\left( {{2^n} - 1} \right)}}

Option A (

\frac{{{{2^{mn}} - 1}}}{{{{2^{mn}}\left( {{{2^n} - 1}} \right)}}}

) is the correct answer because it represents the sum of the given series after simplification using binomial theorem and geometric progression.

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