If `f(x) = tan^(-1)((2^x)/(1 + 2^(2x + 1)))`, then `sum_(r = 0)^(9) f^(r )` is

Options

A. `tan^(-1) (1024)`
B. `tan^(-1)((1023)/(1024))`
C. `tan^(-1)((1023)/(1025))`
D. None of these

Answer

To solve the problem, we need to evaluate the summation of the function \( f(x) = \tan^{-1}\left(\frac{2^x}{1 + 2^{2x + 1}}\right) \) from \( r = 0 \) to \( r = 9 \).

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \tan^{-1}\left(\frac{2^x}{1 + 2^{2x + 1}}\right) \]
We can rewrite the denominator:
\[ 1 + 2^{2x + 1} = 1 + 2 \cdot 2^{2x} = 1 + 2^{2x} \cdot 2 \]
Thus, we have:
\[ f(x) = \tan^{-1}\left(\frac{2^x}{1 + 2 \cdot 2^{2x}}\right) \]

### Step 2: Simplify the expression
Notice that:
\[ f(x) = \tan^{-1}\left(\frac{2^x}{1 + 2^{2x + 1}}\right) = \tan^{-1}\left(\frac{2^x}{1 + 2^{2x} \cdot 2}\right) \]
This can be simplified further:
\[ = \tan^{-1}\left(\frac{2^x}{2^{2x} + 1}\right) \]

### Step 3: Use the identity for the tangent inverse
Using the identity:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
we can express \( f(x) \) in terms of two arctangents:
\[ f(x) = \tan^{-1}(2^x + 1) - \tan^{-1}(2^x) \]

### Step 4: Set up the summation
Now, we need to evaluate:
\[ \sum_{r=0}^{9} f(r) = \sum_{r=0}^{9} \left( \tan^{-1}(2^r + 1) - \tan^{-1}(2^r) \right) \]
This summation can be rewritten using the properties of summation:
\[ = \left( \tan^{-1}(2^0 + 1) - \tan^{-1}(2^0) \right) + \left( \tan^{-1}(2^1 + 1) - \tan^{-1}(2^1) \right) + \ldots + \left( \tan^{-1}(2^9 + 1) - \tan^{-1}(2^9) \right) \]

### Step 5: Observe the cancellation
Notice that this is a telescoping series. Most terms will cancel out:
\[ = \tan^{-1}(2^{10} + 1) - \tan^{-1}(2^0) \]
Calculating the values:
\[ = \tan^{-1}(1024 + 1) - \tan^{-1}(1) = \tan^{-1}(1025) - \frac{\pi}{4} \]

### Step 6: Final evaluation
The final expression can be simplified further, but for the purpose of the problem, we can state:
\[ \sum_{r=0}^{9} f(r) = \tan^{-1}(1025) - \frac{\pi}{4} \]

### Conclusion
Thus, the result of the summation is:
\[ \sum_{r=0}^{9} f(r) = \tan^{-1}(1025) - \frac{\pi}{4} \]