Let x be the `7^(th)` term from the beginning and y be the `7^(th)` term from the end in the expansion of `(3^(1//3)+(1)/(4^(1//3)))^(n`. If `y=12x` then the value of n is :

Options

A. 9
B. 8
C. 10
D. 11

Answer

To solve the problem, we need to find the value of \( n \) such that the 7th term from the beginning \( x \) and the 7th term from the end \( y \) in the expansion of \( (3^{1/3} + \frac{1}{4^{1/3}})^n \) satisfy the equation \( y = 12x \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_k \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_k = \binom{n}{k-1} a^{n-k+1} b^{k-1} \]
Here, \( a = 3^{1/3} \) and \( b = \frac{1}{4^{1/3}} \).

2. **Find the 7th Term from the Beginning**:
The 7th term from the beginning corresponds to \( k = 7 \), so:
\[ x = T_7 = \binom{n}{6} (3^{1/3})^{n-6} \left(\frac{1}{4^{1/3}}\right)^6 \]
Simplifying this gives:
\[ x = \binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{(4^{1/3})^6} = \binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{4^2} = \binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{16} \]

3. **Find the 7th Term from the End**:
The 7th term from the end corresponds to \( k = n - 6 \), so:
\[ y = T_{n-6} = \binom{n}{n-6} (3^{1/3})^6 \left(\frac{1}{4^{1/3}}\right)^{n-6} \]
Simplifying this gives:
\[ y = \binom{n}{6} \cdot (3^{1/3})^6 \cdot \left(\frac{1}{4^{1/3}}\right)^{n-6} = \binom{n}{6} \cdot 3^2 \cdot \frac{1}{(4^{1/3})^{n-6}} = \binom{n}{6} \cdot 9 \cdot \frac{1}{4^{(n-6)/3}} \]

4. **Set Up the Equation**:
According to the problem, we have:
\[ y = 12x \]
Substituting the expressions for \( x \) and \( y \):
\[ \binom{n}{6} \cdot 9 \cdot \frac{1}{4^{(n-6)/3}} = 12 \left(\binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{16}\right) \]

5. **Cancel \( \binom{n}{6} \)**:
Since \( \binom{n}{6} \) is common on both sides, we can cancel it (assuming \( n \geq 6 \)):
\[ 9 \cdot \frac{1}{4^{(n-6)/3}} = 12 \cdot 3^{(n-6)/3} \cdot \frac{1}{16} \]

6. **Simplify the Equation**:
Multiply both sides by \( 16 \):
\[ 144 \cdot \frac{1}{4^{(n-6)/3}} = 12 \cdot 3^{(n-6)/3} \]
Dividing both sides by 12:
\[ 12 \cdot \frac{1}{4^{(n-6)/3}} = 3^{(n-6)/3} \]

7. **Express in Powers**:
Recall that \( 4 = 2^2 \), so:
\[ 12 = 3 \cdot 4 \implies 12 = 3 \cdot (2^2) \implies 12 = 3^{1} \cdot 2^{4} \]
Thus:
\[ 12 \cdot \frac{1}{(2^2)^{(n-6)/3}} = 3^{(n-6)/3} \]

8. **Equate the Powers**:
This leads us to:
\[ 3^{1} \cdot 2^{4 - \frac{2(n-6)}{3}} = 3^{\frac{n-6}{3}} \]
Equating the powers of 3 gives:
\[ 1 = \frac{n-6}{3} \implies n - 6 = 3 \implies n = 9 \]

### Final Answer:
The value of \( n \) is \( 9 \).