Ratio of the `5^(th)` term from the beginning to the `5^(th)` term from the end in the binomial expansion of `(2^(1//3)+(1)/(2(3)^(1//3)))^(10)` is

Options

A. `1:4(16)^(1/3)`
B. `1:2(6)^(1/3)`
C. `2(36)^(1/3):1`
D. `4(36)^((1)/(3)):1`

Answer

To solve the problem, we need to find the ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of \((2^{1/3} + \frac{1}{2(3^{1/3})})^{10}\).

### Step 1: Identify the terms in the binomial expansion
The general term in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = 2^{1/3}\), \(b = \frac{1}{2(3^{1/3})}\), and \(n = 10\).

### Step 2: Find the 5th term from the beginning
The 5th term from the beginning corresponds to \(r = 4\) (since \(T_{r+1}\)):
\[ T_5 = \binom{10}{4} (2^{1/3})^{10-4} \left(\frac{1}{2(3^{1/3})}\right)^4 \]
Calculating this:
\[ T_5 = \binom{10}{4} (2^{1/3})^6 \left(\frac{1}{2(3^{1/3})}\right)^4 \]
\[ = \binom{10}{4} (2^{2}) (2^{-4}) (3^{-4/3}) \]
\[ = \binom{10}{4} \frac{2^{2-4}}{3^{4/3}} = \binom{10}{4} \frac{1}{2^{2} \cdot 3^{4/3}} \]

### Step 3: Find the 5th term from the end
The 5th term from the end corresponds to \(r = 5\) (since \(T_{n-r+1}\)):
\[ T_6 = \binom{10}{5} (2^{1/3})^{10-5} \left(\frac{1}{2(3^{1/3})}\right)^5 \]
Calculating this:
\[ T_6 = \binom{10}{5} (2^{1/3})^5 \left(\frac{1}{2(3^{1/3})}\right)^5 \]
\[ = \binom{10}{5} (2^{5/3}) \left(\frac{1}{2^5 \cdot (3^{1/3})^5}\right) \]
\[ = \binom{10}{5} \frac{2^{5/3}}{2^5 \cdot 3^{5/3}} = \binom{10}{5} \frac{2^{5/3 - 5}}{3^{5/3}} = \binom{10}{5} \frac{2^{-10/3}}{3^{5/3}} \]

### Step 4: Find the ratio of the two terms
Now we need to find the ratio:
\[ \text{Ratio} = \frac{T_5}{T_6} = \frac{\binom{10}{4} \frac{1}{2^{2} \cdot 3^{4/3}}}{\binom{10}{5} \frac{2^{-10/3}}{3^{5/3}}} \]
This simplifies to:
\[ = \frac{\binom{10}{4}}{\binom{10}{5}} \cdot \frac{3^{5/3}}{3^{4/3}} \cdot \frac{2^{-10/3}}{2^{2}} \]
\[ = \frac{\binom{10}{4}}{\binom{10}{5}} \cdot 3^{1} \cdot 2^{-10/3 - 2} \]
\[ = \frac{10! / (4!6!)}{10! / (5!5!)} \cdot 3 \cdot 2^{-16/3} \]
\[ = \frac{5}{6} \cdot 3 \cdot 2^{-16/3} \]
\[ = \frac{15}{6} \cdot 2^{-16/3} = \frac{5}{2} \cdot 2^{-16/3} \]

### Final Step: Simplify the ratio
The final ratio can be expressed as:
\[ = \frac{5}{2^{1 + 16/3}} = \frac{5}{2^{19/3}} \]

### Conclusion
Thus, the ratio of the 5th term from the beginning to the 5th term from the end is:
\[ \frac{5}{2^{19/3}} \]