Question

In the expansion of \[ \left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n , \, n \in \mathbb{N}, \] if the ratio of the 15^th term from the beginning to the 15^th term from the end is \[ \frac{1}{6}, \] then the value of \[ {}^nC_3 \] is:

• A. 4060
• B. 1040
• C. 2300
• D. 4960

Hint:

Use the binomial theorem to find the general term in the expansion.

Answer 1

The Correct Option is C

1. General term in the binomial expansion: \[ T_{r+1} = ^nC_r \left( \sqrt{5} \right)^{n-r} \left( \frac{1}{\sqrt{5}} \right)^r = ^nC_r \left( \sqrt{5} \right)^{n-2r} \]
2. Given ratio of the $15^{th}$ term from the beginning to the $15^{th}$ term from the end: \[ \frac{T_{15}}{T_{n-13}} = \frac{1}{6} \]
3. Express the terms: \[ T_{15} = ^nC_{14} \left( \sqrt{5} \right)^{n-28} \] \[ T_{n-13} = ^nC_{14} \left( \sqrt{5} \right)^{28-n} \]
4. Set up the ratio and solve for n: \[ \frac{^nC_{14} \left( \sqrt{5} \right)^{n-28}}{^nC_{14} \left( \sqrt{5} \right)^{28-n}} = \frac{1}{6} \] \[ \left( \sqrt{5} \right)^{n-56} = \frac{1}{6} \] \[ \left( \sqrt{5} \right)^{n-56} = 6^{-1} \] \[ n - 56 = -1 \implies n = 55 \]5. Calculate $^nC_3$ using the determined value of n: \[ ^nC_3 = ^{55}C_3 = \frac{55 \cdot 54 \cdot 53}{3 \cdot 2 \cdot 1} = 2300 \]Therefore, the correct answer is (3) 2300.