For any positive n, let `f(n)=(4n+sqrt(4n^(2)-1))/(sqrt(2n+1)+sqrt(2n-1))`. Then `((Sigma_(k=1)^(40)f(k))/(100))` is equal to

Answer

To solve the problem, we need to simplify the function \( f(n) \) and then compute the sum \( \sum_{k=1}^{40} f(k) \) divided by 100.

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

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

First, we can rationalize the denominator. The denominator is \( \sqrt{2n + 1} + \sqrt{2n - 1} \). We multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{2n + 1} - \sqrt{2n - 1} \):

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

### Step 2: Simplifying the Denominator

The denominator simplifies as follows:
\[ (\sqrt{2n + 1})^2 - (\sqrt{2n - 1})^2 = (2n + 1) - (2n - 1) = 2 \]

### Step 3: Simplifying the Numerator

Now, we simplify the numerator:
\[ (4n + \sqrt{4n^2 - 1})(\sqrt{2n + 1} - \sqrt{2n - 1}) \]

Expanding this gives:
\[ 4n(\sqrt{2n + 1} - \sqrt{2n - 1}) + \sqrt{4n^2 - 1}(\sqrt{2n + 1} - \sqrt{2n - 1}) \]

### Step 4: Further Simplifying \( f(n) \)

After rationalizing, we can express \( f(n) \) as:
\[ f(n) = \frac{4n(\sqrt{2n + 1} - \sqrt{2n - 1}) + \sqrt{4n^2 - 1}(\sqrt{2n + 1} - \sqrt{2n - 1})}{2} \]

### Step 5: Finding \( \sum_{k=1}^{40} f(k) \)

Now, we need to compute:
\[ \sum_{k=1}^{40} f(k) \]

Using the simplified form of \( f(n) \):
\[ f(n) = \frac{(2n + 1)^{3/2} - (2n - 1)^{3/2}}{2} \]

We can express the sum as:
\[ \sum_{k=1}^{40} f(k) = \sum_{k=1}^{40} \frac{(2k + 1)^{3/2} - (2k - 1)^{3/2}}{2} \]

This is a telescoping series. Most terms will cancel out, leaving us with:
\[ \frac{1}{2} \left( (2 \cdot 40 + 1)^{3/2} - (2 \cdot 1 - 1)^{3/2} \right) \]

Calculating the remaining terms:
\[ = \frac{1}{2} \left( 81^{3/2} - 1^{3/2} \right) = \frac{1}{2} \left( 729 - 1 \right) = \frac{728}{2} = 364 \]

### Step 6: Final Calculation

Now, we divide by 100:
\[ \frac{364}{100} = 3.64 \]

### Final Answer
Thus, the value of \( \frac{\sum_{k=1}^{40} f(k)}{100} \) is \( \boxed{3.64} \).