Let `n` be a positive integer if `1le1gt K len` such that `(sin^(2)nx)/(sin^(2)x)=a_(@)+sum_(1ge i lt klen) a_(1,k) cos 2 (k-1)` for all real number `x` with `x` not an integer multiple of `pi`, then the value of `a_(1,k)` is

Explanation

To solve the given problem, we need to analyze the expression provided and derive the coefficients \( a_{1,k} \). Let's break it down step by step.

### Step 1: Understand the Given Expression
We are given the equation:
\[ \frac{\sin^2(nx)}{\sin^2(x)} = a_0 + \sum_{1 \leq k \leq n} a_{1,k} \cos(2(k-1)x) \]
for all real numbers \( x \) that are not integer multiples of \( \pi \).

### Step 2: Rewrite the Left-Hand Side
Using the identity for sine, we can express \( \sin^2(nx) \) in terms of sine functions:
\[ \sin^2(nx) = \frac{1 - \cos(2nx)}{2} \]
Thus, we can write:
\[ \frac{\sin^2(nx)}{\sin^2(x)} = \frac{1 - \cos(2nx)}{2\sin^2(x)} \]

### Step 3: Simplify the Expression
Now, let's express \( \sin^2(x) \) in terms of cosine:
\[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]
Substituting this back into our expression gives:
\[ \frac{\sin^2(nx)}{\sin^2(x)} = \frac{1 - \cos(2nx)}{1 - \cos(2x)} \]

### Step 4: Expand the Right-Hand Side
The right-hand side is already given in a summation form. We need to find the coefficients \( a_{1,k} \) that match the left-hand side.

### Step 5: Identify the Coefficients
From the expression, we can see that the left-hand side can be expanded in terms of cosines. The term \( \cos(2nx) \) will contribute to the coefficients in the summation on the right-hand side.

### Step 6: Determine \( a_{1,k} \)
By comparing the coefficients of \( \cos(2(k-1)x) \) from both sides, we can conclude that:
\[ a_{1,k} = 2 \quad \text{for all } k = 1, 2, \ldots, n \]

### Final Answer
Thus, the value of \( a_{1,k} \) is:
\[ \boxed{2} \]