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 (2 + 1 2 ) ^ 200 . If (C99 200 K)/(a) = 2 ^ l m n , where m and n are odd

Correct answer: C. (50, 101) In the binomial expansion of (1 + x) ^ 99 , the sum of odd coefficients of x is given by K = C _ 1 + C _ 3 + . + C _ 99 = 2 ^ 99 - 0 2 = 2 ^ 98 Also, a = Middle term in the binomial expansion of (2 + 1 2 ) ^ 200 is T _ (200)/(2) + 1 = T _ 100 + 1 = C100 200 (2) ^ 100 ( 1 2 ) ^ 100 ⇒ T _ 100 + 1 = C100 200 · 2 ^ 50 So, (C99 200 K)/(a) = C99 200 × 2 ^ 98 C100 200 × 2 ^ 50 = (100)/(101) × 2 ^ 48 = (25)/(101) × 2 ^ 50 = ((m)/(n)) 2 ^ l On comparing, we get l = 50, m = 25andn = 101 Therefore, (l, n) ≡ (50, 101) .

JEE Main 202329 JanuaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2023 · 29 January · Evening Shift · Mathematics · Binomial Theorem

Let K be the sum of the coefficients of the odd powers of x in the...

Let

K

be the sum of the coefficients of the odd powers of

x

in the expansion of

{\left(1 + x\right)}^{99}

. Let a be the middle term in the expansion of

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

. If

\dfrac{C99 200 K}{a} = \dfrac{{2}^{l}m}{n}

, where

m and n

are odd numbers, then the ordered pair \left(l, n\right) is equal to:

Options

A $\left(50, 51\right)$
B $\left(51, 99\right)$
C $\left(50, 101\right)$
D $\left(51, 101\right)$

Correct answer

C. $\left(50, 101\right)$

Step-by-step solution

In the binomial expansion of
${\left(1 + x\right)}^{99}$ , the sum of odd coefficients of $x$ is given by

$K = {C}_{1} + {C}_{3} + \ldots . + {C}_{99} = \dfrac{{2}^{99} - 0}{2} = {2}^{98}$

Also, $a =$ Middle term in the binomial expansion of ${\left(2 + \dfrac{1}{\sqrt{2}}\right)}^{200} is$ {T}_{\dfrac{200}{2} + 1} = {T}_{100 + 1} = C100 200 {\left(2\right)}^{100}{\left(\dfrac{1}{\sqrt{2}}\right)}^{100} $$ ⇒ {T}_{100 + 1} = C100 200 \cdot {2}^{50} $So,$ \dfrac{C99 200 K}{a} $$ = \dfrac{C99 200 \times {2}^{98}}{C100 200 \times {2}^{50}} = \dfrac{100}{101} \times {2}^{48} $$ = \dfrac{25}{101} \times {2}^{50} = \left(\dfrac{m}{n}\right){2}^{l} $On comparing, we get$ l = 50, m = 25andn = 101 $Therefore, \left(l, n\right) ≡ \left(50, 101\right)$ .

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