If $n$ is an odd natural number and $^n{C_0}{ ^n}{C_{r + 1}}{ > ^n}{C_{r + 2}} > {....^n}{C_n},$ then $r = $

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Understanding +4 | -1

n

is an odd natural number and

^n{C_0}{ < ^n}{C_1}{ < ^n}{C_2} < {....^n}{C_r}{ > ^n}{C_{r + 1}}{ > ^n}{C_{r + 2}} > {....^n}{C_n},

then

r =

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because when n is odd, the greatest binomial coefficient occurs at the two middle terms, which are

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

and

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

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