` 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, we need to evaluate the summation:

\[ \sum_{k=1}^{n} \tan^{-1}\left(\frac{1}{1+k+k^2}\right) \]

### Step 1: Rewrite the term inside the summation

We start by rewriting the term \(\tan^{-1}\left(\frac{1}{1+k+k^2}\right)\). We can use the identity for the difference of arctangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab} \right) \]

We can express \(\frac{1}{1+k+k^2}\) in terms of \(\tan^{-1}\):

\[ \tan^{-1}\left(\frac{1}{1+k+k^2}\right) = \tan^{-1}(k+1) - \tan^{-1}(k) \]

### Step 2: Substitute into the summation

Now we substitute this expression back into the summation:

\[ \sum_{k=1}^{n} \tan^{-1}\left(\frac{1}{1+k+k^2}\right) = \sum_{k=1}^{n} \left(\tan^{-1}(k+1) - \tan^{-1}(k)\right) \]

### Step 3: Recognize the telescoping series

Notice that this summation is a telescoping series. The terms will cancel out:

\[ = \left(\tan^{-1}(2) - \tan^{-1}(1)\right) + \left(\tan^{-1}(3) - \tan^{-1}(2)\right) + \ldots + \left(\tan^{-1}(n+1) - \tan^{-1}(n)\right) \]

### Step 4: Simplify the result

When we expand this, we see that all intermediate terms cancel:

\[ = \tan^{-1}(n+1) - \tan^{-1}(1) \]

### Step 5: Final result

Thus, the final result of the summation is:

\[ \sum_{k=1}^{n} \tan^{-1}\left(\frac{1}{1+k+k^2}\right) = \tan^{-1}(n+1) - \tan^{-1}(1) \]