If p is nearly equal to q and n `gt` 1 , such that
`((n+1) p+(n-1)q)/((n-1)p+(n +1)q) = ((p)/(q))^(k)` , then the value of k, is

Options

A. n = 2r is a positive integral mulitple of 3
B. `(1)/(n)`
C. n+1
D. `(1)/(n+1)`

Answer

To solve the given problem, we need to find the value of \( k \) in the equation:

\[ \frac{(n+1)p + (n-1)q}{(n-1)p + (n+1)q} = \left(\frac{p}{q}\right)^k \]

where \( p \) is nearly equal to \( q \) and \( n > 1 \).

### Step-by-Step Solution:

1. **Assume \( p \) in terms of \( q \)**:
Since \( p \) is nearly equal to \( q \), we can express \( p \) as:
\[ p = q + h \]
where \( h \) is a very small quantity.

2. **Substitute \( p \) in the LHS**:
Substitute \( p \) in the left-hand side (LHS) of the equation:
\[ \text{LHS} = \frac{(n+1)(q+h) + (n-1)q}{(n-1)(q+h) + (n+1)q} \]

3. **Simplify the LHS**:
Expanding the numerator:
\[ \text{Numerator} = (n+1)q + (n+1)h + (n-1)q = nq + q + (n+1)h = (2n)q + (n+1)h \]
Expanding the denominator:
\[ \text{Denominator} = (n-1)q + (n-1)h + (n+1)q = nq + (n-1)h \]
Thus, the LHS becomes:
\[ \text{LHS} = \frac{(2n)q + (n+1)h}{nq + (n-1)h} \]

4. **Divide numerator and denominator by \( q \)**:
To simplify further, divide both the numerator and denominator by \( q \):
\[ \text{LHS} = \frac{2n + \frac{(n+1)h}{q}}{n + \frac{(n-1)h}{q}} \]

5. **Use the Binomial Theorem**:
Since \( h \) is very small, we can use the Binomial theorem to approximate:
\[ \frac{1 + x}{1 + y} \approx 1 + (x - y) \quad \text{for small } x \text{ and } y \]
Here, let \( x = \frac{(n+1)h}{q} \) and \( y = \frac{(n-1)h}{q} \):
\[ \text{LHS} \approx 1 + \left(\frac{(n+1)h}{q} - \frac{(n-1)h}{q}\right) = 1 + \frac{2h}{q} \]

6. **Equate LHS and RHS**:
The right-hand side (RHS) can be approximated as:
\[ \text{RHS} = \left(1 + \frac{h}{q}\right)^k \approx 1 + k \frac{h}{q} \]
Setting LHS equal to RHS gives:
\[ 1 + \frac{2h}{q} \approx 1 + k \frac{h}{q} \]

7. **Solve for \( k \)**:
By comparing coefficients of \( h \):
\[ \frac{2}{q} = k \frac{1}{q} \]
Thus, we find:
\[ k = 2 \]

### Final Answer:
The value of \( k \) is:
\[ \boxed{2} \]