If the coefficients of `x^(7)` and `x^(6)` in `(2+(x)/(3))^(n)` are equal, then n is

Options

A. 56
B. 55
C. 48
D. 15

Answer

To solve the problem, we need to find the value of \( n \) such that the coefficients of \( x^7 \) and \( x^6 \) in the expansion of \( \left(2 + \frac{x}{3}\right)^n \) are equal.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The coefficient of \( x^r \) in the expansion of \( (a + b)^n \) is given by the formula:
\[ \text{Coefficient of } x^r = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = 2 \) and \( b = \frac{x}{3} \).

2. **Calculate the coefficient of \( x^7 \)**:
For \( r = 7 \):
\[ \text{Coefficient of } x^7 = \binom{n}{7} \cdot 2^{n-7} \cdot \left(\frac{1}{3}\right)^7 \]
Simplifying this gives:
\[ \text{Coefficient of } x^7 = \binom{n}{7} \cdot 2^{n-7} \cdot \frac{1}{2187} \]

3. **Calculate the coefficient of \( x^6 \)**:
For \( r = 6 \):
\[ \text{Coefficient of } x^6 = \binom{n}{6} \cdot 2^{n-6} \cdot \left(\frac{1}{3}\right)^6 \]
Simplifying this gives:
\[ \text{Coefficient of } x^6 = \binom{n}{6} \cdot 2^{n-6} \cdot \frac{1}{729} \]

4. **Set the coefficients equal**:
According to the problem, the coefficients of \( x^7 \) and \( x^6 \) are equal:
\[ \binom{n}{7} \cdot 2^{n-7} \cdot \frac{1}{2187} = \binom{n}{6} \cdot 2^{n-6} \cdot \frac{1}{729} \]

5. **Cross-multiply to eliminate fractions**:
\[ \binom{n}{7} \cdot 2^{n-7} \cdot 729 = \binom{n}{6} \cdot 2^{n-6} \cdot 2187 \]

6. **Simplify the equation**:
Dividing both sides by \( 2^{n-7} \):
\[ \binom{n}{7} \cdot 729 = \binom{n}{6} \cdot 2 \cdot 2187 \]
Since \( 2187 = 3^7 \), we can simplify further:
\[ \binom{n}{7} \cdot 729 = \binom{n}{6} \cdot 4374 \]

7. **Use the relationship between binomial coefficients**:
Recall that:
\[ \binom{n}{7} = \frac{n!}{7!(n-7)!} \quad \text{and} \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]
Thus, we can express \( \binom{n}{7} \) in terms of \( \binom{n}{6} \):
\[ \binom{n}{7} = \frac{n-6}{7} \cdot \binom{n}{6} \]

8. **Substituting back**:
Substitute \( \binom{n}{7} \) into the equation:
\[ \frac{n-6}{7} \cdot \binom{n}{6} \cdot 729 = \binom{n}{6} \cdot 4374 \]
Cancel \( \binom{n}{6} \) (assuming \( n \geq 6 \)):
\[ \frac{n-6}{7} \cdot 729 = 4374 \]

9. **Solve for \( n \)**:
Multiply both sides by 7:
\[ n - 6 = \frac{4374 \cdot 7}{729} \]
Simplifying gives:
\[ n - 6 = 42 \quad \Rightarrow \quad n = 48 \]

### Final Answer:
Thus, the value of \( n \) is \( \boxed{48} \).