JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Analyzing +4 | -1

Sum of the series

\sum\limits_{k = 0}^n {^n{C_k}{{\left( { - 1} \right)}^k}\frac{1}{{{a_k}}}}

where

a_k=\sum\limits_{{i=0}}^k{ }^k C_i b_i

where

b_i=\sum\limits_{{j=0}}^i{ }^i C_j\left(\frac{-2}{3}\right)^j

View answer and explanation

Correct option: C

\frac{1}{4^n}

Option C,

\frac{{1}}{{4^n}}

, is correct because after simplifying the series, it becomes a binomial expansion equal to

(\frac{{1}}{{4}})^n

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