If $n - {\;^1}{C_r} = \left( {{k^2} - 3} \right){\;^n}{C_{r + 1}},$ then $k \in $

$\left( { - \infty ,\; - 2} \right)$

$\left[ { - \sqrt 3 ,\;\sqrt 3 } \right]$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

n - {\;^1}{C_r} = \left( {{k^2} - 3} \right){\;^n}{C_{r + 1}},

then

k \in

View answer and explanation

Correct option: D

\left( {\sqrt 3 ,2} \right]

Option D (

\left( {\sqrt 3 ,2} \right]

) is correct because it satisfies the condition

0 < k^2 - 3 \le 1

derived from the properties of binomial coefficients.

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