` sum _(k=1)^(n) tan^(-1). 1/(1+k+k^(2))` is equal to

Options

A. `tan^(-1) (n+1)`
B. `tan^(-1)(n+2)`
C. `tan^(-1).n/(n+2)`
D. `pi/2`

Answer

To solve the problem \( \sum_{k=1}^{n} \tan^{-1} \left( \frac{1}{1+k+k^2} \right) \), we will use properties of the inverse tangent function.

### Step-by-step Solution:

1. **Rewrite the term inside the summation**:
We start with the term \( \tan^{-1} \left( \frac{1}{1+k+k^2} \right) \). We can use the identity:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right) \]
to express this in a more manageable form.

2. **Identify \( a \) and \( b \)**:
We can express \( \frac{1}{1+k+k^2} \) in terms of \( k \) and \( k+1 \):
\[ \tan^{-1} \left( \frac{1}{1+k+k^2} \right) = \tan^{-1} (k+1) - \tan^{-1} (k) \]
This is because:
\[ \frac{(k+1) - k}{1 + k(k+1)} = \frac{1}{1+k+k^2} \]

3. **Rewrite the summation**:
Now, we can rewrite the summation:
\[ \sum_{k=1}^{n} \left( \tan^{-1} (k+1) - \tan^{-1} (k) \right) \]

4. **Recognize the telescoping nature**:
Notice that this is a telescoping series. When we expand it, we have:
\[ (\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + \ldots + (\tan^{-1}(n+1) - \tan^{-1}(n)) \]
Most terms will cancel out.

5. **Simplify the result**:
After cancellation, we are left with:
\[ \tan^{-1}(n+1) - \tan^{-1}(1) \]

6. **Final result**:
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), the final result is:
\[ \tan^{-1}(n+1) - \frac{\pi}{4} \]

### Conclusion:
Thus, the sum \( \sum_{k=1}^{n} \tan^{-1} \left( \frac{1}{1+k+k^2} \right) \) is equal to:
\[ \tan^{-1}(n+1) - \frac{\pi}{4} \]