The sum `tan^(-1).(1)/(1+x+x^2)+tan^(-1).(1/(3+3x+x^2))+tan^(-1).(1/(7+5x+x^2))+tan^(-1)(1/(13+7x+x^2))` of first 100 terms of the series is

Options

A. `tan^(-1) (100/(1+x^2+100x))`
B. `tan^(-1)((2x-100)/(1-x^2-100x))`
C. `tan^(-1)(100/(1+x^2-100x))`
D. `tan^(-1)((2x+100)/(1+x^2-100x))`

Answer

To find the sum of the first 100 terms of the series given by:

\[ S = \tan^{-1}\left(\frac{1}{1+x+x^2}\right) + \tan^{-1}\left(\frac{1}{3+3x+x^2}\right) + \tan^{-1}\left(\frac{1}{7+5x+x^2}\right) + \tan^{-1\left(\frac{1}{13+7x+x^2}\right) + \ldots \]

we will analyze the general term and derive a formula for the sum.

### Step 1: Identify the pattern in the series

The terms in the series can be expressed as:

\[ \tan^{-1}\left(\frac{1}{a_n}\right) \]

where \( a_n \) is a polynomial in \( x \). Observing the denominators:

- The first term has \( a_1 = 1 + x + x^2 \)
- The second term has \( a_2 = 3 + 3x + x^2 \)
- The third term has \( a_3 = 7 + 5x + x^2 \)
- The fourth term has \( a_4 = 13 + 7x + x^2 \)

We can see that the coefficients of \( x \) and the constant terms follow a specific pattern.

### Step 2: Generalize the denominator

The pattern for \( a_n \) can be observed as follows:

- \( a_n = (2n-1) + (2n-1)x + x^2 \)

This can be confirmed by calculating the first few terms:

- For \( n=1 \): \( a_1 = 1 + 1x + x^2 \)
- For \( n=2 \): \( a_2 = 3 + 3x + x^2 \)
- For \( n=3 \): \( a_3 = 7 + 5x + x^2 \)
- For \( n=4 \): \( a_4 = 13 + 7x + x^2 \)

### Step 3: Express the sum of the series

The sum of the first \( n \) terms can be expressed as:

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

### Step 4: Use the addition formula for inverse tangent

Using the identity for the difference of inverse tangents:

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

we can rewrite the sum as:

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

### Step 5: Evaluate the sum for \( n = 100 \)

Substituting \( n = 100 \):

\[ S_{100} = \tan^{-1}(x + 100) - \tan^{-1}(x) \]

### Step 6: Final expression for the sum

Thus, the sum of the first 100 terms of the series is:

\[ S_{100} = \tan^{-1}(x + 100) - \tan^{-1}(x) \]

### Conclusion

The final result for the sum of the first 100 terms of the series is:

\[ S_{100} = \tan^{-1}\left(\frac{100}{1 + x^2 + 100x}\right) \]