The sum of series `sec^(-1)sqrt(2)+sec^(-1)(sqrt(10))/3+sec^(-1)(sqrt(50))/7++sec^(-1)sqrt(((n^2+1)(n^2-2n+2))/((n^2-n+1)^2))` is `tan^(-1)1` (b) `n` `tan^(-1)(n+1)` (d) `tan^(-1)(n-1)`

Options

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

Explanation

To solve the problem, we need to evaluate the sum of the series given by:

\[ S = \sec^{-1}(\sqrt{2}) + \frac{\sec^{-1}(\sqrt{10})}{3} + \frac{\sec^{-1}(\sqrt{50})}{7} + \cdots + \sec^{-1}\left(\sqrt{\frac{(n^2+1)(n^2-2n+2)}{(n^2-n+1)^2}}\right) \]

Let's denote the general term of the series as \( T_n \):

\[ T_n = \sec^{-1}\left(\sqrt{\frac{(n^2+1)(n^2-2n+2)}{(n^2-n+1)^2}}\right) \]

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

We need to simplify the expression inside the secant inverse function:

\[ \sqrt{\frac{(n^2+1)(n^2-2n+2)}{(n^2-n+1)^2}} \]

### Step 2: Rewrite the Expression

The expression can be rewritten as:

\[ \sqrt{\frac{(n^2 + 1)(n^2 - 2n + 2)}{(n^2 - n + 1)^2}} = \frac{\sqrt{(n^2 + 1)(n^2 - 2n + 2)}}{n^2 - n + 1} \]

### Step 3: Using the Identity for Secant Inverse

We know that:

\[ \sec^{-1}(x) = \tan^{-1}\left(\sqrt{x^2 - 1}\right) \]

Thus, we can express \( T_n \) in terms of \( \tan^{-1} \):

\[ T_n = \tan^{-1}\left(\sqrt{\frac{(n^2 + 1)(n^2 - 2n + 2)}{(n^2 - n + 1)^2} - 1}\right) \]

### Step 4: Finding the Sum of the Series

The series can be expressed as:

\[ S = T_1 + T_2 + \cdots + T_n \]

Using the properties of inverse tangent, we can combine the terms. The series can be simplified to:

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

### Step 5: Final Result

Thus, the sum of the series is:

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

### Conclusion

The final answer is:

\[ \boxed{\tan^{-1}(n + 1)} \]