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 step by step, we start with the function defined as:

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

We need to calculate:

\[ \frac{\sum_{k=1}^{40} f(k)}{100} \]

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

We can rewrite \( f(n) \) as follows:

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

Notice that \( \sqrt{4n^2 - 1} = \sqrt{(2n - 1)(2n + 1)} \). Thus, we can express \( f(n) \) as:

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

### Step 2: Rationalizing the Denominator

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

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

The denominator simplifies to:

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

Thus, we have:

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

### Step 3: Further Simplifying \( f(n) \)

Expanding the numerator:

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

This expression can be complex, but we can observe that it leads to a telescoping series when we compute \( \sum_{k=1}^{40} f(k) \).

### Step 4: Evaluating the Sum

When we compute \( \sum_{k=1}^{40} f(k) \), we find that many terms cancel out. Specifically, we can express \( f(k) \) in terms of \( k \):

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

Thus, the sum becomes:

\[ \sum_{k=1}^{40} f(k) = \frac{1}{2} \left( (81^{3/2} - 1^{3/2}) \right) \]

### Step 5: Calculating the Final Value

Calculating \( 81^{3/2} \):

\[ 81^{3/2} = (9^2)^{3/2} = 9^3 = 729 \]

So, we have:

\[ \sum_{k=1}^{40} f(k) = \frac{1}{2} (729 - 1) = \frac{728}{2} = 364 \]

Finally, we compute:

\[ \frac{\sum_{k=1}^{40} f(k)}{100} = \frac{364}{100} = 3.64 \]

### Final Answer

Thus, the value of \( \frac{\sum_{k=1}^{40} f(k)}{100} \) is:

\[ \boxed{3.64} \]