Let `f (n)=(4n + sqrt(4n ^(2) +1))/( sqrt(2n +1 )+sqrt(2n-1)),n in N` then the remainder when `f (1) + f (2) + f (3) + ..... + f (60)` is divided by 9 is.

Answer

To solve the problem, we need to evaluate the function \( f(n) \) defined as:

\[ f(n) = \frac{4n + \sqrt{4n^2 + 1}}{\sqrt{2n + 1} + \sqrt{2n - 1}} \]

and find the remainder when \( f(1) + f(2) + f(3) + \ldots + f(60) \) is divided by 9.

### Step 1: Simplify \( f(n) \)

To simplify \( f(n) \), we multiply the numerator and the denominator by \( \sqrt{2n + 1} - \sqrt{2n - 1} \):

\[ f(n) = \frac{(4n + \sqrt{4n^2 + 1})(\sqrt{2n + 1} - \sqrt{2n - 1})}{(\sqrt{2n + 1} + \sqrt{2n - 1})(\sqrt{2n + 1} - \sqrt{2n - 1})} \]

The denominator simplifies to:

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

Thus, we have:

\[ f(n) = \frac{(4n + \sqrt{4n^2 + 1})(\sqrt{2n + 1} - \sqrt{2n - 1})}{2} \]

### Step 2: Further Simplification

Now, we can simplify the numerator:

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

This leads to:

\[ f(n) = \frac{(2n + 1)^{3/2} - (2n - 1)^{3/2}}{2} \]

### Step 3: Calculate \( f(1) + f(2) + \ldots + f(60) \)

Now we can express the sum:

\[ f(1) + f(2) + \ldots + f(60) = \frac{1}{2} \left( (3^{3/2} - 1^{3/2}) + (5^{3/2} - 3^{3/2}) + (7^{3/2} - 5^{3/2}) + \ldots + (121^{3/2} - 119^{3/2}) \right) \]

Notice that this is a telescoping series. Most terms will cancel out, leaving us with:

\[ = \frac{1}{2} \left( 121^{3/2} - 1^{3/2} \right) \]

### Step 4: Calculate \( 121^{3/2} \)

Calculating \( 121^{3/2} \):

\[ 121^{3/2} = (11^2)^{3/2} = 11^3 = 1331 \]

Thus, we have:

\[ f(1) + f(2) + \ldots + f(60) = \frac{1}{2} (1331 - 1) = \frac{1330}{2} = 665 \]

### Step 5: Find the Remainder When Divided by 9

Now we need to find the remainder when 665 is divided by 9:

\[ 665 \div 9 = 73 \quad \text{(quotient)} \]
\[ 73 \times 9 = 657 \]
\[ 665 - 657 = 8 \]

Thus, the remainder when \( f(1) + f(2) + \ldots + f(60) \) is divided by 9 is:

\[ \boxed{8} \]