`sum_(m=1)^(n) tan^(-1) ((2m)/(m^(4) + m^(2) + 2))` is equal to

Options

A. `tan^(-1) ((n^(2) + n)/(n^(2) + n + 2))`
B. `tan^(-1) ((n^(2) -n)/(n^(2) - n + 2))`
C. `tan^(-1) ((n^(2) + n + 2)/(n^(2) + n))`
D. none of these

Explanation

To solve the problem \( \sum_{m=1}^{n} \tan^{-1} \left( \frac{2m}{m^4 + m^2 + 2} \right) \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression inside the summation:
\[ \tan^{-1} \left( \frac{2m}{m^4 + m^2 + 2} \right) \]
We can rewrite the denominator:
\[ m^4 + m^2 + 2 = (m^2 + 1)^2 + 1 \]
Thus, we have:
\[ \tan^{-1} \left( \frac{2m}{(m^2 + 1)^2 + 1} \right) \]

### Step 2: Use the identity for the difference of arctangents
We can express the term as a difference of two arctangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a - b}{1 + ab} \right) \]
We can set:
\[ a = m^2 + m + 1 \quad \text{and} \quad b = m^2 - m + 1 \]
Then we have:
\[ \tan^{-1}(m^2 + m + 1) - \tan^{-1}(m^2 - m + 1) \]

### Step 3: Substitute back into the summation
Now, we can rewrite the summation:
\[ \sum_{m=1}^{n} \left( \tan^{-1}(m^2 + m + 1) - \tan^{-1}(m^2 - m + 1) \right) \]

### Step 4: Recognize the telescoping nature
This summation is telescoping:
\[ = \left( \tan^{-1}(2) - \tan^{-1}(1) \right) + \left( \tan^{-1}(7) - \tan^{-1}(3) \right) + \ldots + \left( \tan^{-1}(n^2 + n + 1) - \tan^{-1}(n^2 - n + 1) \right) \]
Most terms will cancel out, leaving us with:
\[ = \tan^{-1}(n^2 + n + 1) - \tan^{-1}(1) \]

### Step 5: Final simplification
Using the identity for the difference of arctangents:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \]
We can substitute \( x = n^2 + n + 1 \) and \( y = 1 \):
\[ = \tan^{-1} \left( \frac{(n^2 + n + 1) - 1}{1 + (n^2 + n + 1) \cdot 1} \right) \]
This simplifies to:
\[ = \tan^{-1} \left( \frac{n^2 + n}{2 + n^2 + n} \right) \]

### Conclusion
Thus, the final result is:
\[ \sum_{m=1}^{n} \tan^{-1} \left( \frac{2m}{m^4 + m^2 + 2} \right) = \tan^{-1} \left( \frac{n^2 + n}{2 + n^2 + n} \right) \]