Find the sum of each of the following series :(i) `tan^-1(1/(x^2+x+1))+tan^-1 (1/(x^2+3x+3))+tan^-1(1/(x^2+5X+7))+tan^-1(1/x^2+7x+13))`......upto n.

Explanation

To find the sum of the series

\[ S_n = \tan^{-1}\left(\frac{1}{x^2+x+1}\right) + \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) + \tan^{-1}\left(\frac{1}{x^2+5x+7}\right) + \tan^{-1\left(\frac{1}{x^2+7x+13}\right) + \ldots \text{ up to } n \]

we can use the identity for the difference of inverse tangents:

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

### Step 1: Rewrite the terms in the series

The terms can be expressed as:

- \(\tan^{-1}\left(\frac{1}{x^2 + 2kx + (k^2 - k + 1)}\right)\) for \(k = 1, 2, 3, \ldots, n\).

### Step 2: Identify the pattern

Notice that the denominators can be rewritten:

1. \(x^2 + 1x + 1\) can be expressed as \(x^2 + 2(1)x + (1^2 - 1 + 1)\)
2. \(x^2 + 3x + 3\) can be expressed as \(x^2 + 2(2)x + (2^2 - 2 + 1)\)
3. \(x^2 + 5x + 7\) can be expressed as \(x^2 + 2(3)x + (3^2 - 3 + 1)\)
4. \(x^2 + 7x + 13\) can be expressed as \(x^2 + 2(4)x + (4^2 - 4 + 1)\)

This suggests a general form for the \(k\)-th term.

### Step 3: Apply the identity

Using the identity, we can pair the terms:

\[ S_n = \left(\tan^{-1}(A) - \tan^{-1}(B)\right) + \left(\tan^{-1}(B) - \tan^{-1}(C)\right) + \ldots \]

This leads to a telescoping series where most terms cancel out.

### Step 4: Simplify the expression

After cancellation, we are left with:

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

### Step 5: Final expression

Thus, the sum of the series is:

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