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 sum \( \sum_{r=0}^{9} f(r) \) where \( f(x) = \tan^{-1}\left(\frac{2^x}{1 + 2^{2x + 1}}\right) \).

### Step-by-Step Solution:

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:
\[ 1 + 2^{2x + 1} = 1 + 2 \cdot 2^{2x} = 1 + 2 \cdot (2^x)^2 \]
Thus, we have:
\[ f(x) = \tan^{-1}\left(\frac{2^x}{1 + 2 \cdot (2^x)^2}\right) \]

2. **Use the Identity for Tangent Inverse**:
We can use the identity:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Here, we can express \( f(x) \) in a different form:
\[ f(x) = \tan^{-1}(2^{x}) - \tan^{-1}(2^{x + 1}) \]

3. **Sum the Function**:
Now we compute the sum:
\[ \sum_{r=0}^{9} f(r) = \sum_{r=0}^{9} \left( \tan^{-1}(2^r) - \tan^{-1}(2^{r + 1}) \right) \]
This is a telescoping series. Most terms will cancel out:
\[ = \left( \tan^{-1}(2^0) - \tan^{-1}(2^{10}) \right) \]
Since \( 2^0 = 1 \) and \( 2^{10} = 1024 \), we have:
\[ = \tan^{-1}(1) - \tan^{-1}(1024) \]

4. **Evaluate the Inverse Tangent**:
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]
Therefore:
\[ \sum_{r=0}^{9} f(r) = \frac{\pi}{4} - \tan^{-1}(1024) \]

5. **Final Expression**:
The final result can be expressed as:
\[ \sum_{r=0}^{9} f(r) = \tan^{-1}\left(\frac{1}{1024}\right) \]

### Final Result:
Thus, the required sum is:
\[ \sum_{r=0}^{9} f(r) = \tan^{-1}(1023) - \tan^{-1}(1024) \]