JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Analyzing +4 | -1

Let

K

be the sum of the coefficients of the odd powers of

x

in the expansion of

(1+x)^{99}

. Let

a

be the middle term in the expansion of

\left(2+\frac{1}{\sqrt{2}}\right)^{200}

. If

\frac{{ }^{200} C_{99} K}{a}=\frac{2^{\prime} m}{n}

, where

m

and

n

are odd numbers, then the ordered pair

(I, n)

is equal to

View answer and explanation

Correct option: C

(50,101)

Option C (

(50,101)

) is correct because it is obtained by calculating the sum of coefficients of odd powers, the middle term, and then simplifying the given 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