Let f be defined on the natural numbers as follow: f(1)=1 and for `n gt 1, f(n)=f[f(n-1)]+f[n-f(n-1)]`, the value of `1/(30)sum_(r=1)^(20)f(r)` is

Answer

To solve the given problem, we will first compute the values of the function \( f(n) \) for \( n = 1 \) to \( n = 20 \) based on the recursive definition provided.

### Step-by-Step Solution:

1. **Define the function**:
- We know \( f(1) = 1 \).
- For \( n > 1 \), the function is defined as:
\[ f(n) = f(f(n-1)) + f(n - f(n-1)) \]

2. **Calculate \( f(2) \)**:
- Using the recursive formula:
\[ f(2) = f(f(1)) + f(2 - f(1)) = f(1) + f(1) = 1 + 1 = 2 \]

3. **Calculate \( f(3) \)**:
- Using the recursive formula:
\[ f(3) = f(f(2)) + f(3 - f(2)) = f(2) + f(3 - 2) = f(2) + f(1) = 2 + 1 = 3 \]

4. **Calculate \( f(4) \)**:
- Using the recursive formula:
\[ f(4) = f(f(3)) + f(4 - f(3)) = f(3) + f(4 - 3) = f(3) + f(1) = 3 + 1 = 4 \]

5. **Calculate \( f(5) \)**:
- Using the recursive formula:
\[ f(5) = f(f(4)) + f(5 - f(4)) = f(4) + f(5 - 4) = f(4) + f(1) = 4 + 1 = 5 \]

6. **Continue this pattern**:
- It appears that \( f(n) = n \) for \( n = 1, 2, 3, 4, 5 \).
- By induction, we can show that \( f(n) = n \) holds for all \( n \) in the natural numbers.

7. **Sum the values**:
- Since \( f(r) = r \) for \( r = 1 \) to \( 20 \):
\[ \sum_{r=1}^{20} f(r) = \sum_{r=1}^{20} r = 1 + 2 + 3 + \ldots + 20 \]
- The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- Therefore:
\[ \sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210 \]

8. **Calculate the final value**:
- We need to find:
\[ \frac{1}{30} \sum_{r=1}^{20} f(r) = \frac{1}{30} \times 210 = 7 \]

### Final Answer:
\[ \frac{1}{30} \sum_{r=1}^{20} f(r) = 7 \]