JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

C_r

stands for

{ }^n C_r

, then the sum of the series

\frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}\left[C_0^2-2 C_1^2+3 C_2^2 \ldots \ldots+(-1)^n(n+1) C_n^2\right]

where '

{n}

' is an even positive integer is equal to

View answer and explanation

Correct option: C

(-1)^{n / 2}(n+2)

Option C (

(-1)^{n / 2}(n+2)

) is correct because after simplifying the series, the sum equals this expression.

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