If `2f(x) = f(2x), 3f(x) = f(3x), 4f(x) = f(4x)…….` etc. then `f(1) + f(2) + f(3) + ……..+ f(n)` equals to where `f(1) = 1` :

Options

A. a.`n(n + 1)`
B. b. `0`
C. c. `(n(n+1))/(2)`
D. d. `n!`

Answer

To solve the problem, we start with the given equations:

1. \( 2f(x) = f(2x) \)
2. \( 3f(x) = f(3x) \)
3. \( 4f(x) = f(4x) \)
4. and so on...

We are also given that \( f(1) = 1 \).

### Step 1: Finding the general form of \( f(x) \)

From the equation \( 2f(x) = f(2x) \), we can deduce the behavior of the function \( f \). Let's evaluate it at \( x = 1 \):

\[ 2f(1) = f(2) \]

Substituting \( f(1) = 1 \):

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

Now, let's evaluate \( f(3) \) using \( 3f(1) = f(3) \):

\[ 3f(1) = f(3) \implies 3 \cdot 1 = f(3) \implies f(3) = 3 \]

Next, we can find \( f(4) \) using \( 4f(1) = f(4) \):

\[ 4f(1) = f(4) \implies 4 \cdot 1 = f(4) \implies f(4) = 4 \]

Continuing this pattern, we can see that:

\[ f(n) = n \quad \text{for } n = 1, 2, 3, 4, \ldots \]

### Step 2: Summing \( f(1) + f(2) + f(3) + \ldots + f(n) \)

Now we need to find the sum:

\[ f(1) + f(2) + f(3) + \ldots + f(n) \]

Since we have established that \( f(n) = n \), we can rewrite the sum as:

\[ 1 + 2 + 3 + \ldots + n \]

### Step 3: Using the formula for the sum of the first \( n \) natural numbers

The formula for the sum of the first \( n \) natural numbers is:

\[ \text{Sum} = \frac{n(n + 1)}{2} \]

### Final Result

Thus, we conclude that:

\[ f(1) + f(2) + f(3) + \ldots + f(n) = \frac{n(n + 1)}{2} \]