Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3} \mathrm{x}^{\frac{1}{3}}+\frac{1}{2 \mathrm{x}^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Binomial Theorem

Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of

(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x ^{\frac{2}{3}}})^{18}

. Then

(\frac{n}{m})^{\frac{1}{3}}

is :

Answer: (D)

General term of

(\frac{x ^{1/3}}{3} + \frac{1}{2 x ^{2/3}})^{18}

T_{r + 1} = (r 18) (\frac{x ^{1/3}}{3})^{18 - r} (\frac{1}{2 x ^{2/3}})^{r} = (r 18) \frac{1}{3 ^{18 - r} 2 ^{r}} x^{(18 - r) /3 - 2 r /3} .

The 7th term corresponds to

r = 6

; its coefficient is

m = (6 18) \frac{1}{3 ^{12} 2 ^{6}}

.

The 13th term corresponds to

r = 12

; its coefficient is

n = (12 18) \frac{1}{3 ^{6} 2 ^{12}}

.

Since

(12 18) = (6 18)

, these binomial coefficients cancel in the ratio:

\frac{n}{m} = \frac{3 ^{12} 2 ^{6}}{3 ^{6} 2 ^{12}} = 3^{6} \cdot 2^{- 6} = (\frac{3}{2})^{6} .

(\frac{n}{m})^{1/3} = (\frac{3}{2})^{2} = \frac{9}{4} .

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option