The value of ${ }^n C_3+{ }^n C_7+{ }^n C_{11}+\ldots$ is

$\frac{1}{4}\left\{2^{n+1}-2^{n / 2} \sin \frac{n \pi}{4}\right\}$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The value of

{ }^n C_3+{ }^n C_7+{ }^n C_{11}+\ldots

View answer and explanation

Correct option: A

\frac{1}{2}\left\{2^{n-1}-2^{n / 2} \sin \frac{n \pi}{4}\right\}

Option A (

\frac{{1}}{{2}}\left{2^{{n-1}}-2^{{n / 2}} \sin \frac{{n \pi}}{{4}}\right}

) is correct because by using the binomial theorem and complex numbers, the given series can be expressed and simplified to

\frac{{1}}{{2}}\left{2^{{n-1}}-2^{{n / 2}} \sin \frac{{n \pi}}{{4}}\right}

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