If `f(x+y+z)=f(x)+f(y)+f(z)` with `f(1)=1` and `f(2)=2` and `x,y, z epsilonR` the value of `lim_(xtooo)sum_(r=1)^(n)((4r)f(3r))/(n^(3))` is ……….

Answer

To solve the problem step by step, we start with the given functional equation and the limit we need to evaluate.

### Step 1: Understanding the functional equation
We have the functional equation:
\[ f(x+y+z) = f(x) + f(y) + f(z) \]
This suggests that \( f \) is a linear function. We also know:
\[ f(1) = 1 \]
\[ f(2) = 2 \]

### Step 2: Finding the form of \( f(x) \)
From the functional equation, we can deduce that \( f \) is likely of the form \( f(x) = kx \) for some constant \( k \).

Using the given values:
- For \( f(1) = 1 \):
\[ k \cdot 1 = 1 \implies k = 1 \]
- For \( f(2) = 2 \):
\[ k \cdot 2 = 2 \implies k = 1 \]

Thus, we conclude:
\[ f(x) = x \]

### Step 3: Substitute \( f(3r) \) into the limit
Now, we need to evaluate the limit:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{4r f(3r)}{n^3} \]
Since \( f(3r) = 3r \), we can substitute this into the expression:
\[ = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{4r \cdot 3r}{n^3} \]
\[ = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{12r^2}{n^3} \]

### Step 4: Simplifying the summation
The summation can be expressed as:
\[ = \lim_{n \to \infty} \frac{12}{n^3} \sum_{r=1}^{n} r^2 \]

Using the formula for the sum of squares:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]

Substituting this into our limit:
\[ = \lim_{n \to \infty} \frac{12}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \]
\[ = \lim_{n \to \infty} \frac{2n(n+1)(2n+1)}{n^3} \]

### Step 5: Simplifying further
Now we simplify the expression:
\[ = \lim_{n \to \infty} \frac{2(2n^3 + 3n^2 + n)}{n^3} \]
\[ = \lim_{n \to \infty} 2 \left( 2 + \frac{3}{n} + \frac{1}{n^2} \right) \]

### Step 6: Evaluating the limit
As \( n \to \infty \):
\[ \frac{3}{n} \to 0 \quad \text{and} \quad \frac{1}{n^2} \to 0 \]
Thus, we have:
\[ = 2(2 + 0 + 0) = 4 \]

### Final Answer
The value of the limit is:
\[ \boxed{4} \]