In the expansion of `(1+x)^n ,` 7th and 8th terms are equal. Find the value of `(7//x+6)^2` .

Explanation

To solve the problem, we need to find the value of \((\frac{7}{x} + 6)^2\) given that the 7th and 8th terms in the expansion of \((1 + x)^n\) are equal.

### Step-by-step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \((1 + x)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} x^r \]
where \(T_{r+1}\) is the \((r+1)\)th term.

2. **Write the 7th and 8th Terms**:
The 7th term corresponds to \(r = 6\):
\[ T_7 = \binom{n}{6} x^6 \]
The 8th term corresponds to \(r = 7\):
\[ T_8 = \binom{n}{7} x^7 \]

3. **Set the 7th and 8th Terms Equal**:
According to the problem, the 7th and 8th terms are equal:
\[ \binom{n}{6} x^6 = \binom{n}{7} x^7 \]

4. **Simplify the Equation**:
We can simplify this equation by dividing both sides by \(x^6\) (assuming \(x \neq 0\)):
\[ \binom{n}{6} = \binom{n}{7} x \]

5. **Express \(x\)**:
Rearranging gives us:
\[ x = \frac{\binom{n}{6}}{\binom{n}{7}} \]
We know that:
\[ \binom{n}{7} = \frac{n-6}{7} \binom{n}{6} \]
Therefore:
\[ x = \frac{\binom{n}{6}}{\frac{n-6}{7} \binom{n}{6}} = \frac{7}{n-6} \]

6. **Substitute \(x\) into the Expression**:
Now we need to find \((\frac{7}{x} + 6)^2\):
First, calculate \(\frac{7}{x}\):
\[ \frac{7}{x} = \frac{7}{\frac{7}{n-6}} = n - 6 \]

7. **Calculate the Final Expression**:
Now substitute this into the expression:
\[ \frac{7}{x} + 6 = (n - 6) + 6 = n \]
Therefore:
\[ \left(\frac{7}{x} + 6\right)^2 = n^2 \]

### Final Answer:
The value of \((\frac{7}{x} + 6)^2\) is:
\[ \boxed{n^2} \]