In the expansion of \(\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}, \mathrm{n} \in \mathrm{N}\), if the ratio of \(15^{\text {th }}\) term from the beginning to the \(15^{\text {th }}\) term from the end is \(\frac{1}{6}\), then the value of \({ }^{n} C_{3}\) is:

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Binomial Theorem

In the expansion of

(32 + \frac{1}{3 3})^{n}, n \in N

, if the ratio of

1 5^{th}

term from the beginning to the

1 5^{th}

term from the end is

\frac{1}{6}

, then the value of

^{n} C_{3}

is:

Answer: (C)

With general term

T_{r + 1} = (r n) (2^{1/3})^{n - r} (3^{- 1/3})^{r}

, the

15

th term from the start uses

r = 14

and the

15

th from the end uses

r = n - 14

, and their binomial coefficients are equal. The ratio reduces to

(6^{1/3})^{n - 28} = 6^{(n - 28) /3}

, and setting it equal to

\frac{1}{6} = 6^{- 1}

gives

\frac{n - 28}{3} = - 1

, so

n = 25

. Then

(3 25) = \frac{25 \cdot 24 \cdot 23}{6} = 2300

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option