Suppose `f:R rarr(0,oo)` be a differentiable function such that `5f(x+y)=f(x)*f(y),AA x,y in R`. If `f(3)=320,` then `sum_(n=0)^(5)f(n)` is equal to :

Options

A. `6575`
B. `6825`
C. `6875`
D. `6525`

Answer

To solve the problem, we start with the functional equation given:

\[ 5f(x+y) = f(x) \cdot f(y) \]

for all \( x, y \in \mathbb{R} \), and we know that \( f(3) = 320 \).

### Step 1: Find \( f(0) \)

Let \( x = 0 \) and \( y = 0 \):

\[ 5f(0+0) = f(0) \cdot f(0) \implies 5f(0) = f(0)^2 \]

This can be rearranged to:

\[ f(0)^2 - 5f(0) = 0 \implies f(0)(f(0) - 5) = 0 \]

Thus, \( f(0) = 0 \) or \( f(0) = 5 \). Since \( f: \mathbb{R} \to (0, \infty) \), we have:

\[ f(0) = 5 \]

### Step 2: Find \( f(1) \)

Next, let \( y = 1 \):

\[ 5f(x+1) = f(x) \cdot f(1) \]

Now, let’s set \( x = 1 \):

\[ 5f(2) = f(1) \cdot f(1) \implies 5f(2) = f(1)^2 \]

### Step 3: Find \( f(2) \)

Let \( x = 2 \) and \( y = 1 \):

\[ 5f(3) = f(2) \cdot f(1) \]

We know \( f(3) = 320 \):

\[ 5 \cdot 320 = f(2) \cdot f(1) \implies 1600 = f(2) \cdot f(1) \]

### Step 4: Find \( f(1) \)

From the equation \( 5f(2) = f(1)^2 \) and \( 1600 = f(2) \cdot f(1) \), we can express \( f(2) \) in terms of \( f(1) \):

\[ f(2) = \frac{f(1)^2}{5} \]

Substituting this into the second equation:

\[ 1600 = \left(\frac{f(1)^2}{5}\right) \cdot f(1) \implies 1600 = \frac{f(1)^3}{5} \]

Multiplying both sides by 5:

\[ 8000 = f(1)^3 \implies f(1) = 20 \]

### Step 5: Find \( f(2) \)

Now substituting \( f(1) \) back to find \( f(2) \):

\[ f(2) = \frac{20^2}{5} = \frac{400}{5} = 80 \]

### Step 6: Find \( f(4) \) and \( f(5) \)

Using the functional equation again:

\[ 5f(3) = f(2) \cdot f(1) \implies 5 \cdot 320 = 80 \cdot 20 \]

This is consistent, so we can find \( f(4) \) and \( f(5) \):

Using \( x = 3 \) and \( y = 1 \):

\[ 5f(4) = f(3) \cdot f(1) \implies 5f(4) = 320 \cdot 20 = 6400 \implies f(4) = \frac{6400}{5} = 1280 \]

Using \( x = 4 \) and \( y = 1 \):

\[ 5f(5) = f(4) \cdot f(1) \implies 5f(5) = 1280 \cdot 20 = 25600 \implies f(5) = \frac{25600}{5} = 5120 \]

### Step 7: Calculate \( \sum_{n=0}^{5} f(n) \)

Now we can calculate:

\[ \sum_{n=0}^{5} f(n) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5) \]

Substituting the values we found:

\[ = 5 + 20 + 80 + 320 + 1280 + 5120 \]

Calculating this gives:

\[ = 5 + 20 = 25 \]
\[ = 25 + 80 = 105 \]
\[ = 105 + 320 = 425 \]
\[ = 425 + 1280 = 1705 \]
\[ = 1705 + 5120 = 6825 \]

Thus, the final answer is:

\[ \sum_{n=0}^{5} f(n) = 6825 \]