The value of `cot(sum_(n=1)^50 tan^-1((1)/(1+n+n^2)))` is equal to

Options

A. `25/26`
B. `3/25`
C. `26/25`
D. `3/26`

Answer

To solve the problem, we need to find the value of \( \cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right) \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression inside the cotangent:
\[ \sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right) \]

2. **Rewriting the Argument of the Arctangent**:
We can rewrite the term \( \frac{1}{1+n+n^2} \) as follows:
\[ \frac{1}{1+n+n^2} = \frac{1}{(n+1) + n^2} \]
We can also express this as:
\[ \tan^{-1}\left(\frac{1}{1+n+n^2}\right) = \tan^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right) \]
This can be recognized as:
\[ \tan^{-1}(a - b) = \tan^{-1}(a) - \tan^{-1}(b) \]
where \( a = n+1 \) and \( b = n \).

3. **Using the Arctangent Difference Identity**:
Therefore, we can express the sum as:
\[ \sum_{n=1}^{50} \left( \tan^{-1}(n+1) - \tan^{-1}(n) \right) \]
This is a telescoping series.

4. **Evaluating the Telescoping Series**:
The telescoping nature means that most terms will cancel out:
\[ \tan^{-1}(51) - \tan^{-1}(1) \]

5. **Calculating the Result**:
We know that:
\[ \tan^{-1}(51) - \tan^{-1}(1) = \tan^{-1}\left(\frac{51 - 1}{1 + 51 \cdot 1}\right) = \tan^{-1}\left(\frac{50}{52}\right) = \tan^{-1}\left(\frac{25}{26}\right) \]

6. **Finding the Cotangent**:
Now, we need to find:
\[ \cot\left(\tan^{-1}\left(\frac{25}{26}\right)\right) \]
The cotangent of an angle whose tangent is \( \frac{25}{26} \) is:
\[ \cot\left(\tan^{-1}\left(\frac{25}{26}\right)\right) = \frac{1}{\tan\left(\tan^{-1}\left(\frac{25}{26}\right)\right)} = \frac{26}{25} \]

### Final Answer:
Thus, the value of \( \cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right) \) is:
\[ \frac{26}{25} \]