$\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + . \ldots {\rm{is\;equal\;to}}$

$\frac{{{2^{n + 1}}}}{{n + 1}}$

$\frac{{{2^{n + 1}} - 1}}{{n + 1}}$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + . \ldots \mathrm{is\;equal\;to}

View answer and explanation

Correct option: C

\frac{{{2^n}}}{{n + 1}}

Option C (

\frac{{{2^n}}}{{n + 1}}

) is correct because the given series can be expressed as an integral, and the result simplifies to

\frac{{{2^n}}}{{n + 1}}

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