The sum of the series
`(1)/(sqrt(1)+sqrt(2))+(1)/(sqrt(2)+sqrt(3))+(1)/(sqrt(3)+sqrt(4))+ . . . . .+(1)/(sqrt(n^(2)-1)+sqrt(n^(2)))` equals

Options

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

Answer

To find the sum of the series

\[ S = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \ldots + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}, \]

we will simplify each term in the series.

### Step 1: Rationalizing the Denominator
We start with the first term:

\[ \frac{1}{\sqrt{k} + \sqrt{k+1}}. \]

To rationalize the denominator, we multiply the numerator and denominator by \(\sqrt{k} - \sqrt{k+1}\):

\[ \frac{1}{\sqrt{k} + \sqrt{k+1}} \cdot \frac{\sqrt{k} - \sqrt{k+1}}{\sqrt{k} - \sqrt{k+1}} = \frac{\sqrt{k} - \sqrt{k+1}}{(\sqrt{k} + \sqrt{k+1})(\sqrt{k} - \sqrt{k+1})}. \]

Using the difference of squares in the denominator:

\[ (\sqrt{k})^2 - (\sqrt{k+1})^2 = k - (k+1) = -1, \]

we get:

\[ \frac{\sqrt{k} - \sqrt{k+1}}{-1} = -(\sqrt{k} - \sqrt{k+1}) = \sqrt{k+1} - \sqrt{k}. \]

### Step 2: Expressing the Series
Now, we can rewrite the series \(S\) as:

\[ S = \left(\sqrt{2} - \sqrt{1}\right) + \left(\sqrt{3} - \sqrt{2}\right) + \left(\sqrt{4} - \sqrt{3}\right) + \ldots + \left(\sqrt{n^2} - \sqrt{n^2 - 1}\right). \]

### Step 3: Telescoping Series
Notice that this is a telescoping series. Most terms will cancel out:

\[ S = \sqrt{n^2} - \sqrt{1} = n - 1. \]

### Conclusion
Thus, the sum of the series is:

\[ S = n - 1. \]