किसी घनात्मक पूर्णांक n के लिए यदि,
`(1)/(sqrt(4)+sqrt(5))+(1)/(sqrt(5)+sqrt(6)) +(1)/(sqrt(6)+sqrt(7)) +..... + (1)/(sqrt(n)+sqrt(n+1))=5` है तो n का मान क्या होगा?

Options

A. 45
B. 46
C. 47
D. 48

Answer

To solve the equation

\[ \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \ldots + \frac{1}{\sqrt{n} + \sqrt{n+1}} = 5, \]

we will simplify each term in the sum and find the value of \( n \).

### Step 1: Simplify each term

We can simplify each term of the form

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

To simplify, we can multiply the numerator and denominator by the conjugate of the denominator:

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

The denominator simplifies as follows:

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

Thus, we have:

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

### Step 2: Rewrite the sum

Now, we can rewrite the entire sum:

\[ \sum_{k=4}^{n} \left( \sqrt{k+1} - \sqrt{k} \right). \]

This is a telescoping series. When we expand it, we get:

\[ (\sqrt{5} - \sqrt{4}) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + \ldots + (\sqrt{n+1} - \sqrt{n}). \]

### Step 3: Simplify the telescoping series

In a telescoping series, most terms cancel out:

\[ \sqrt{n+1} - \sqrt{4}. \]

### Step 4: Set the equation

Now we set this equal to 5:

\[ \sqrt{n+1} - 2 = 5. \]

### Step 5: Solve for \( n \)

Adding 2 to both sides gives:

\[ \sqrt{n+1} = 7. \]

Now, squaring both sides results in:

\[ n + 1 = 49. \]

Subtracting 1 from both sides, we find:

\[ n = 48. \]

### Final Answer

Thus, the value of \( n \) is

\[ \boxed{48}. \]