JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Analyzing +4 | -1

{C_\mathrm{r}}

stands for

{\;^n}{C_r}

, the sum of the given series

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

where

n

is an even positive integer, is

View answer and explanation

Correct option: D

{\left( { - 1} \right)^{n/2}}\left( {n + 2} \right)

Option D (

{{\left( { - 1} \right)}^{n/2}}\left( {n + 2} \right)

) is correct because simplifying the given series using properties of binomial coefficients leads to the result

{{\left( { - 1} \right)}^{n/2}}\left( {n + 2} \right)

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