The value of the expression `cot^(-1) (1/2) + cot^(-1) (9/2) + cot^(-1) (25/2) + cot^(-1) (49/2)` upto + .......n terms is

Options

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

Answer

To solve the expression \( \cot^{-1} \left( \frac{1}{2} \right) + \cot^{-1} \left( \frac{9}{2} \right) + \cot^{-1} \left( \frac{25}{2} \right) + \cot^{-1} \left( \frac{49}{2} \right) \) up to \( n \) terms, we can use some properties of inverse trigonometric functions.

### Step 1: Rewrite the cotangent inverse
We can use the property that \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \). Therefore, we can rewrite each term in the expression:
\[ \cot^{-1} \left( \frac{1}{2} \right) = \tan^{-1}(2), \quad \cot^{-1} \left( \frac{9}{2} \right) = \tan^{-1}\left(\frac{2}{9}\right), \quad \cot^{-1} \left( \frac{25}{2} \right) = \tan^{-1}\left(\frac{2}{25}\right), \quad \cot^{-1} \left( \frac{49}{2} \right) = \tan^{-1}\left(\frac{2}{49}\right) \]

### Step 2: Identify the pattern
Notice that the terms can be expressed in a general form:
\[ \cot^{-1} \left( \frac{(2k-1)^2}{2} \right) \text{ for } k = 1, 2, 3, \ldots, n \]
This means we can express the summation as:
\[ \sum_{k=1}^{n} \cot^{-1} \left( \frac{(2k-1)^2}{2} \right) \]

### Step 3: Use the addition formula
Using the addition formula for inverse tangent:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \]
we can express the sum as:
\[ \tan^{-1}(2) - \tan^{-1}(0) + \tan^{-1}(4) - \tan^{-1}(2) + \tan^{-1}(6) - \tan^{-1}(4) + \ldots + \tan^{-1}(2n) - \tan^{-1}(2n-2) \]

### Step 4: Cancel terms
Notice that in this telescoping series, all terms cancel except for the last term:
\[ \tan^{-1}(2n) - \tan^{-1}(0) \]
Since \( \tan^{-1}(0) = 0 \), we have:
\[ \tan^{-1}(2n) \]

### Step 5: Final expression
Thus, the value of the expression up to \( n \) terms is:
\[ \tan^{-1}(2n) \]

### Conclusion
Therefore, the final answer is:
\[ \boxed{\tan^{-1}(2n)} \]