The value of ` sum_(m=1)^(n) "tan"^(-1)((2m)/(m^(4) + m^(2) +2 )) ` is :

Options

A. ` tan^(-1)(n^(2) +n) `
B. ` "tan"^(-1)((n^(2)+n +1)/(n^(2) +n + 2))`
C. ` "tan"^(-1)((n)/(n+1))`
D. ` "tan"^(-1)((n^(2) +n)/(n^(2) + n + 2)) `

Answer

To solve the problem, we need to evaluate the summation:

\[ S = \sum_{m=1}^{n} \tan^{-1}\left(\frac{2m}{m^4 + m^2 + 2}\right) \]

### Step 1: Simplifying the Argument of the Inverse Tangent

First, we simplify the expression inside the summation:

\[ \tan^{-1}\left(\frac{2m}{m^4 + m^2 + 2}\right) \]

We can rewrite the denominator:

\[ m^4 + m^2 + 2 = m^4 + 2 + m^2 = (m^2 + 1)^2 + 1 - m^2 \]

This gives us:

\[ \tan^{-1}\left(\frac{2m}{(m^2 + 1)^2 + 1 - m^2}\right) \]

### Step 2: Using the Addition Formula for Inverse Tangent

We can express the argument in a form suitable for the inverse tangent subtraction formula:

\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy} \]

Let:

\[ x = m^2 + m + 1 \quad \text{and} \quad y = m^2 - m + 1 \]

Then:

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

This simplifies to:

\[ \tan^{-1}\left(\frac{2m}{1 + (m^2 + 1)^2 - m^2}\right) \]

### Step 3: Rewriting the Summation

Now we can rewrite the summation:

\[ S = \sum_{m=1}^{n} \left(\tan^{-1}(m^2 + m + 1) - \tan^{-1}(m^2 - m + 1)\right) \]

### Step 4: Telescoping Series

This summation is telescoping:

\[ S = \left[\tan^{-1}(2) - \tan^{-1}(1)\right] + \left[\tan^{-1}(3) - \tan^{-1}(2)\right] + \ldots + \left[\tan^{-1}(n^2 + n + 1) - \tan^{-1}(n^2 - n + 1)\right] \]

The intermediate terms cancel out, leaving us with:

\[ S = \tan^{-1}(n^2 + n + 1) - \tan^{-1}(1) \]

### Step 5: Final Result

Since \(\tan^{-1}(1) = \frac{\pi}{4}\), we have:

\[ S = \tan^{-1}(n^2 + n + 1) - \frac{\pi}{4} \]

Thus, the final value of the summation is:

\[ S = \tan^{-1}(n^2 + n + 1) - \frac{\pi}{4} \]