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 can simplify each term in the series.

### Step 1: Simplifying Each Term

Each term can be rewritten using the identity:

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

This means we can express each term in the series as:

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

### Step 2: Rewriting the Series

Now, we can rewrite the entire series using this simplification:

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

### Step 3: Observing the Telescoping Nature

Notice that this series is telescoping. Most terms will cancel out:

- The \(-\sqrt{1}\) from the first term remains.
- The last term is \(\sqrt{n^2}\).

Thus, we have:

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

### Step 4: Conclusion

Therefore, the sum of the series is:

\[ \boxed{n - 1}. \]

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