In the expansion of `(2+1/(3x))^(n)`, the cofficient of ` x^(-7) and x^(-8) ` are equal to

Options

A. 51
B. 52
C. 55
D. 56

Answer

To solve the problem, we need to find the value of \( n \) such that the coefficients of \( x^{-7} \) and \( x^{-8} \) in the expansion of \( (2 + \frac{1}{3x})^n \) are equal.

### Step-by-Step Solution:

1. **Write the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case, \( a = 2 \) and \( b = \frac{1}{3x} \). Thus, the general term becomes:
\[ T_{r+1} = \binom{n}{r} \cdot 2^{n-r} \cdot \left(\frac{1}{3x}\right)^r = \binom{n}{r} \cdot 2^{n-r} \cdot \frac{1}{3^r} \cdot x^{-r} \]

2. **Coefficients of \( x^{-7} \) and \( x^{-8} \)**:
- For \( x^{-7} \), set \( r = 7 \):
\[ T_8 = \binom{n}{7} \cdot 2^{n-7} \cdot \frac{1}{3^7} \]
- For \( x^{-8} \), set \( r = 8 \):
\[ T_9 = \binom{n}{8} \cdot 2^{n-8} \cdot \frac{1}{3^8} \]

3. **Set the Coefficients Equal**:
We need to equate the coefficients of \( x^{-7} \) and \( x^{-8} \):
\[ \binom{n}{7} \cdot 2^{n-7} \cdot \frac{1}{3^7} = \binom{n}{8} \cdot 2^{n-8} \cdot \frac{1}{3^8} \]

4. **Simplify the Equation**:
Rearranging gives:
\[ \frac{\binom{n}{7}}{\binom{n}{8}} = \frac{1}{3} \cdot \frac{2^{n-7}}{2^{n-8}} \]
Simplifying further, we have:
\[ \frac{\binom{n}{7}}{\binom{n}{8}} = \frac{2}{3} \]
Using the property of binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \Rightarrow \frac{\binom{n}{7}}{\binom{n}{8}} = \frac{n-7}{8} \]
Thus, we can write:
\[ \frac{n-7}{8} = \frac{2}{3} \]

5. **Cross Multiply**:
Cross multiplying gives:
\[ 3(n-7) = 16 \]
Expanding this results in:
\[ 3n - 21 = 16 \]

6. **Solve for \( n \)**:
Adding 21 to both sides:
\[ 3n = 37 \]
Dividing by 3:
\[ n = \frac{37}{3} \]

### Final Answer:
The value of \( n \) is \( \frac{37}{3} \).