The sum of n terms of the series
`(1)/(sqrt(1) + sqrt3) + 1/(sqrt3 + sqrt5) + 1/(sqrt5 +sqrt7)` + … is :

Options

A. a)`sqrt(2n+1)`
B. b)`1/2 sqrt(2n+1)`
C. c)`sqrt(2n-1)`
D. d)`1/2 {sqrt(2n+1) -1}`

Answer

To find the sum of the first \( n \) terms of the series

\[ S_n = \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \ldots + \frac{1}{\sqrt{2n-1} + \sqrt{2n+1}}, \]

we can start by rewriting each term in a more manageable form.

### Step 1: Rewrite Each Term
Each term can be expressed as:

\[ \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} \quad \text{for } k = 1, 2, \ldots, n. \]

### Step 2: Rationalize the Denominator
To simplify each term, we can multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} \cdot \frac{\sqrt{2k+1} - \sqrt{2k-1}}{\sqrt{2k+1} - \sqrt{2k-1}} = \frac{\sqrt{2k+1} - \sqrt{2k-1}}{(\sqrt{2k-1} + \sqrt{2k+1})(\sqrt{2k+1} - \sqrt{2k-1})}. \]

### Step 3: Simplify the Denominator
The denominator simplifies to:

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

Thus, each term becomes:

\[ \frac{\sqrt{2k+1} - \sqrt{2k-1}}{2}. \]

### Step 4: Write the Sum of the Series
Now, we can express the sum of the first \( n \) terms as:

\[ S_n = \sum_{k=1}^{n} \frac{\sqrt{2k+1} - \sqrt{2k-1}}{2}. \]

### Step 5: Factor Out the Constant
Factoring out \( \frac{1}{2} \):

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

### Step 6: Recognize the Telescoping Nature
Notice that this is a telescoping series. Most terms will cancel out:

\[ S_n = \frac{1}{2} \left( \sqrt{2n+1} - \sqrt{1} \right). \]

### Step 7: Final Simplification
Thus, we have:

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

### Final Answer
The sum of the first \( n \) terms of the series is:

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