Sum of series
`sum_(r=1)^(n)sin^(-1)[(2r+1)/(r(r+1)(sqrt(r^(2)+2r)+sqrt(r^(2)-1)))]`

Options

A. `pi/2 - sin^(-1) ((1)/(n+1))`
B. `cos^(-1) ((1)/(n+1))`
C. `pi/2 -cos^(-1) ((1)/(n+1))`
D. None of these

Answer

To solve the given series

\[ S = \sum_{r=1}^{n} \sin^{-1}\left(\frac{2r + 1}{r(r + 1)(\sqrt{r^2 + 2r} + \sqrt{r^2 - 1})}\right), \]

we will analyze the term inside the summation step by step.

### Step 1: Simplifying the Argument of the Inverse Sine
We start with the expression inside the sine inverse:

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

We can simplify the denominator:

\[ \sqrt{r^2 + 2r} = \sqrt{(r + 1)^2} = r + 1, \]
and
\[ \sqrt{r^2 - 1} = \sqrt{(r - 1)(r + 1)}. \]

Thus, we rewrite the expression as:

\[ \frac{2r + 1}{r(r + 1)(r + 1 + \sqrt{(r - 1)(r + 1})}. \]

### Step 2: Rationalizing the Denominator
Next, we rationalize the denominator. We multiply the numerator and denominator by the conjugate of the denominator:

\[ \sqrt{r^2 + 2r} - \sqrt{r^2 - 1}. \]

This gives us:

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

The denominator simplifies to:

\[ (r + 1)^2 - (r^2 - 1) = 2r + 2. \]

### Step 3: Simplifying Further
The expression now looks like:

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

### Step 4: Using the Identity for Sine Inverse
We can use the identity for the difference of sine inverses:

\[ \sin^{-1}(x) - \sin^{-1}(y) = \sin^{-1}\left(\frac{x - y}{\sqrt{1 - x^2}\sqrt{1 - y^2}}\right). \]

Let \(x = \frac{1}{r}\) and \(y = \frac{1}{r + 1}\). Thus, we can express the series as:

\[ S = \sum_{r=1}^{n} \left(\sin^{-1}\left(\frac{1}{r}\right) - \sin^{-1}\left(\frac{1}{r + 1}\right)\right). \]

### Step 5: Telescoping Series
This is a telescoping series. Most terms cancel out, leaving us with:

\[ S = \sin^{-1}(1) - \sin^{-1}\left(\frac{1}{n + 1}\right). \]

### Step 6: Final Calculation
Since \(\sin^{-1}(1) = \frac{\pi}{2}\), we have:

\[ S = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{n + 1}\right). \]

### Conclusion
Thus, the final result for the sum of the series is:

\[ \boxed{S = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{n + 1}\right)}. \]