In `(3 3+1/(3 3))^n` if the ratio of 7th term from the beginning to the 7th term from the end is 1/6, then find the value of `ndot`

Options

A. 3
B. 5
C. 7
D. 9

Answer

To solve the problem, we need to find the value of \( n \) such that the ratio of the 7th term from the beginning to the 7th term from the end of the binomial expansion of \( \left( 3 + \frac{1}{3} \right)^n \) is \( \frac{1}{6} \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{(r-1)} \]
For our case, \( a = 3 \) and \( b = \frac{1}{3} \).

2. **Calculate the 7th Term from the Beginning**:
The 7th term \( T_7 \) can be calculated as follows:
\[ T_7 = \binom{n}{7-1} \cdot 3^{n-6} \cdot \left(\frac{1}{3}\right)^{6} = \binom{n}{6} \cdot 3^{n-6} \cdot \frac{1}{729} \]
Simplifying this gives:
\[ T_7 = \binom{n}{6} \cdot \frac{3^{n-6}}{729} \]

3. **Calculate the 7th Term from the End**:
The 7th term from the end corresponds to the \( (n-6) \)th term:
\[ T_{n-5} = \binom{n}{n-5} \cdot 3^{n-(n-5)} \cdot \left(\frac{1}{3}\right)^{n-5} = \binom{n}{n-5} \cdot 3^{5} \cdot \left(\frac{1}{3}\right)^{n-5} \]
This simplifies to:
\[ T_{n-5} = \binom{n}{5} \cdot 3^{5} \cdot \frac{1}{3^{n-5}} = \binom{n}{5} \cdot \frac{3^{5}}{3^{n-5}} = \binom{n}{5} \cdot 3^{10-n} \]

4. **Set Up the Ratio**:
We are given that the ratio of the 7th term from the beginning to the 7th term from the end is \( \frac{1}{6} \):
\[ \frac{T_7}{T_{n-5}} = \frac{1}{6} \]
Substituting the expressions we found:
\[ \frac{\binom{n}{6} \cdot \frac{3^{n-6}}{729}}{\binom{n}{5} \cdot 3^{10-n}} = \frac{1}{6} \]

5. **Simplifying the Ratio**:
Rearranging gives:
\[ \frac{\binom{n}{6}}{\binom{n}{5}} \cdot \frac{3^{n-6}}{3^{10-n} \cdot 729} = \frac{1}{6} \]
We know that \( \frac{\binom{n}{6}}{\binom{n}{5}} = \frac{n-5}{6} \), so we substitute that in:
\[ \frac{n-5}{6} \cdot \frac{3^{n-6}}{3^{10-n} \cdot 729} = \frac{1}{6} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ (n-5) \cdot 3^{n-6} = 729 \cdot 3^{10-n} \]
Simplifying further:
\[ (n-5) \cdot 3^{n-6} = 3^6 \cdot 3^{10-n} \]
This leads to:
\[ (n-5) \cdot 3^{n-6} = 3^{16-n} \]

7. **Equating the Powers of 3**:
Since the bases are the same, we can equate the exponents:
\[ n - 6 = 16 - n \]
Solving for \( n \):
\[ 2n = 22 \implies n = 11 \]

### Final Answer:
The value of \( n \) is \( 11 \).