If `S_n=cot^-1(3)+cot^-1(7)+cot^-1(13)+cot^-1(21)+.....`, `n` terms, then

Options

A. `S_(10) = tan^(-1).(5)/(6)`
B. `S_(oo) = (pi)/(4)`
C. `S_(6) = sin^(-1).(4)/(5)`
D. `S_(20) = cot^(-1) 1.1`

Explanation

To solve the problem, we need to evaluate the series \( S_n = \cot^{-1}(3) + \cot^{-1}(7) + \cot^{-1}(13) + \cot^{-1}(21) + \ldots \) up to \( n \) terms.

### Step-by-Step Solution:

1. **Identify the general term**:
The terms in the series can be expressed as:
\[ t_r = \cot^{-1}(1 + r^2 + r) \]
for \( r = 1, 2, 3, \ldots, n \).

2. **Convert cotangent to tangent**:
Using the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \), we rewrite the general term:
\[ t_r = \tan^{-1}\left(\frac{1}{1 + r^2 + r}\right) \]

3. **Simplify the expression**:
We can simplify \( \frac{1}{1 + r^2 + r} \) as follows:
\[ 1 + r^2 + r = (r + 1)(r + 1) - r = (r + 1)^2 - r \]
Thus,
\[ t_r = \tan^{-1}\left(\frac{1}{(r + 1)(r + 1) - r}\right) \]

4. **Use the tangent subtraction formula**:
We can express \( t_r \) using the formula for the difference of two arctangents:
\[ t_r = \tan^{-1}(r + 1) - \tan^{-1}(r) \]

5. **Sum the series**:
The sum \( S_n \) can be expressed as:
\[ S_n = \sum_{r=1}^{n} t_r = \sum_{r=1}^{n} \left( \tan^{-1}(r + 1) - \tan^{-1}(r) \right) \]
This is a telescoping series, which means that most terms will cancel out:
\[ S_n = \tan^{-1}(n + 1) - \tan^{-1}(1) \]

6. **Evaluate specific cases**:
- For \( n = 10 \):
\[ S_{10} = \tan^{-1}(11) - \tan^{-1}(1) \]
- For \( n \to \infty \):
\[ S_{\infty} = \tan^{-1}(\infty) - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]
- For \( n = 6 \):
\[ S_6 = \tan^{-1}(7) - \tan^{-1}(1) \]
- For \( n = 20 \):
\[ S_{20} = \tan^{-1}(21) - \tan^{-1}(1) \]

### Conclusion:
After evaluating the specific cases, we can determine which options are correct based on the values obtained.