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 $15^{th}$ 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: \[ \frac{^nC_{14} \left( \sqrt{5} \right)^{n-28}}{^nC_{14} \left( \sqrt{5} \right)^{28-n}} = \frac{1}{6} \] \[ \left( \sqrt{5} \right)^{n-56} = \frac{1}{6} \] \[ \left( \sqrt{5} \right)^{n-56} = 6^{-1} \] \[ n - 56 = -1 \implies n = 55 \] 5. Calculate $^nC_3$: \[ ^nC_3 = ^{55}C_3 = \frac{55 \cdot 54 \cdot 53}{3 \cdot 2 \cdot 1} = 2300 \] Therefore, the correct answer is (3) 2300.

Answer 2

We expand \( \left( \sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}} \right)^{n} \) and are told that the ratio of the 15th term from the beginning to the 15th term from the end is \( \dfrac{1}{6} \). We must find \( \binom{n}{3} \).

Concept Used:

The \((r+1)\)th term of \((a+b)^n\) is \( T_{r+1}=\binom{n}{r} a^{\,n-r} b^{\,r} \). Also, the \(k\)th term from the end equals the \((n+2-k)\)th term from the beginning.

Step-by-Step Solution:

Step 1: Write the 15th term from the beginning \((r=14)\):

\[ T_{\text{beg}}=\binom{n}{14}\left(\sqrt[3]{2}\right)^{n-14}\left(\frac{1}{\sqrt[3]{3}}\right)^{14} =\binom{n}{14}\,2^{\frac{n-14}{3}}\,3^{-\frac{14}{3}}. \]

Step 2: Write the 15th term from the end, which is the \((n-13)\)th from the beginning \((r=n-14)\):

\[ T_{\text{end}}=\binom{n}{14}\left(\sqrt[3]{2}\right)^{14}\left(\frac{1}{\sqrt[3]{3}}\right)^{n-14} =\binom{n}{14}\,2^{\frac{14}{3}}\,3^{-\frac{n-14}{3}}. \]

Step 3: Form the given ratio and solve for \(n\):

\[ \frac{T_{\text{beg}}}{T_{\text{end}}} =\frac{2^{\frac{n-14}{3}}\,3^{-\frac{14}{3}}}{2^{\frac{14}{3}}\,3^{-\frac{n-14}{3}}} =2^{\frac{n-28}{3}}\,3^{\frac{n-28}{3}} =\left(2\cdot 3\right)^{\frac{n-28}{3}} =6^{\frac{n-28}{3}}. \] Given \( \dfrac{T_{\text{beg}}}{T_{\text{end}}}=\dfrac{1}{6}=6^{-1} \), hence \[ 6^{\frac{n-28}{3}}=6^{-1}\ \Rightarrow\ \frac{n-28}{3}=-1 \ \Rightarrow\ n=25. \]

Step 4: Compute \( \binom{n}{3} \) for \( n=25 \):

\[ \binom{25}{3}=\frac{25\cdot 24\cdot 23}{3\cdot 2\cdot 1} =25\cdot 4\cdot 23=2300. \]

Final Computation & Result

The required value is 2300.