Let K be the sum of the coefficients of the odd powers of x in the expansion of $(1+\mathrm{x})^{99}$. Let a be the middle term in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\frac{{ }^{200} \mathrm{C}_{99} \mathrm{~K}}{\mathrm{a}}=\frac{2^{\ell} \mathrm{m}}{\mathrm{n}}$, where m and n are odd numbers, then the ordered pair ( $\ell, \mathrm{n}$ ) is equal to :

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Binomial Theorem

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 + \frac{1}{2})^{200}

. If

\frac{^{200} C _{99} K}{a} = \frac{2 ^{ℓ} m}{n}

, where m and n are odd numbers, then the ordered pair (

ℓ, n

) is equal to :

Answer: (C)

The sum of the coefficients of the odd powers of

x

(1 + x)^{99}

K = 2^{98}

. The middle term of

(2 + \frac{1}{2})^{200}

a = T_{101} = (100 200) 2^{100} (\frac{1}{2})^{100} = (100 200) 2^{50}

. Hence

\frac{( 99 200 ) K}{a} = \frac{( 99 200 )}{( 100 200 )} \cdot \frac{2 ^{98}}{2 ^{50}} = \frac{100}{101} \cdot 2^{48} = \frac{25 \cdot 2 ^{50}}{101}

. Writing this as

\frac{2 ^{ℓ} m}{n}

with

m, n

odd gives

ℓ = 50

and

n = 101

, so the ordered pair is

(50, 101)

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option