If `sum _(r =1) ^(n) (cos (2rx))/(sin (r x + (pi)/(4)))= (2 sin (50x)sin ((pi)/(4)- (101)/(2)))/(sin ((x)/(2)))` then value of n is `"_________"`

Answer

To solve the equation

\[ \sum_{r=1}^{n} \frac{\cos(2rx)}{\sin\left(rx + \frac{\pi}{4}\right)} = \frac{2 \sin(50x) \sin\left(\frac{\pi}{4} - \frac{101}{2}\right)}{\sin\left(\frac{x}{2}\right)}, \]

we will follow these steps:

### Step 1: Rewrite the left-hand side
We start with the left-hand side:

\[ \sum_{r=1}^{n} \frac{\cos(2rx)}{\sin\left(rx + \frac{\pi}{4}\right)}. \]

To simplify this, we can multiply the numerator and denominator by \(2 \cos\left(rx + \frac{\pi}{4}\right)\):

\[ \sum_{r=1}^{n} \frac{2 \cos(2rx) \cos\left(rx + \frac{\pi}{4}\right)}{2 \cos\left(rx + \frac{\pi}{4}\right) \sin\left(rx + \frac{\pi}{4}\right)}. \]

Using the identity \(2 \sin A \cos A = \sin(2A)\), we can rewrite the numerator:

\[ \sum_{r=1}^{n} \frac{\sin(2rx)}{\sin\left(rx + \frac{\pi}{4}\right)}. \]

### Step 2: Use the sine addition formula
Using the sine addition formula, we can express \(\sin\left(rx + \frac{\pi}{4}\right)\):

\[ \sin\left(rx + \frac{\pi}{4}\right) = \sin(rx)\cos\left(\frac{\pi}{4}\right) + \cos(rx)\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}(\sin(rx) + \cos(rx)). \]

### Step 3: Simplify the sum
Now, we can express the sum in terms of sine and cosine:

\[ \sum_{r=1}^{n} \frac{\sin(2rx)}{\frac{\sqrt{2}}{2}(\sin(rx) + \cos(rx))} = \frac{2}{\sqrt{2}} \sum_{r=1}^{n} \frac{\sin(2rx)}{\sin(rx) + \cos(rx)}. \]

### Step 4: Compare with the right-hand side
Now we can compare this with the right-hand side:

\[ \frac{2 \sin(50x) \sin\left(\frac{\pi}{4} - \frac{101}{2}\right)}{\sin\left(\frac{x}{2}\right)}. \]

### Step 5: Equate coefficients
From the comparison, we can see that:

\[ \frac{n}{2} = 50 \implies n = 100. \]

Thus, the value of \(n\) is

\[ \boxed{100}. \]