JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

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Application +4 | -1

\mathrm{C}_0, \mathrm{C}_1, \mathrm{C}_2 \ldots \ldots \mathrm{C}_{\mathrm{n}}

are binomial coefficients in the expansion of

(1+\mathrm{x})^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}

, then

\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots \ldots+\frac{C_b}{n+1}

is equal to -

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because integrating the binomial expansion and then using specific values for x allows us to isolate the desired sum.

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