Let f be a function from the set of positive integers to the set of real number such that f(1)=1 and `sum_(r=1)^(n)rf(r)=n(n+1)f(n), forall n ge 2` the value of 2126 f(1063) is ………….. .

Answer

To solve the problem, we need to find the value of \( 2126 \cdot f(1063) \) given the function \( f \) defined by the conditions provided. Let's break down the steps:

### Step 1: Understand the given conditions
We know that:
1. \( f(1) = 1 \)
2. The equation \( \sum_{r=1}^{n} r f(r) = n(n+1) f(n) \) holds for all \( n \geq 2 \).

### Step 2: Calculate \( f(2) \)
Substituting \( n = 2 \) into the equation:
\[ \sum_{r=1}^{2} r f(r) = 2(2+1) f(2) \]
This expands to:
\[ 1 \cdot f(1) + 2 \cdot f(2) = 6 f(2) \]
Substituting \( f(1) = 1 \):
\[ 1 + 2 f(2) = 6 f(2) \]
Rearranging gives:
\[ 1 = 6 f(2) - 2 f(2) \implies 1 = 4 f(2) \implies f(2) = \frac{1}{4} \]

### Step 3: Calculate \( f(3) \)
Now, substituting \( n = 3 \):
\[ \sum_{r=1}^{3} r f(r) = 3(3+1) f(3) \]
This expands to:
\[ f(1) + 2 f(2) + 3 f(3) = 12 f(3) \]
Substituting \( f(1) = 1 \) and \( f(2) = \frac{1}{4} \):
\[ 1 + 2 \cdot \frac{1}{4} + 3 f(3) = 12 f(3) \]
This simplifies to:
\[ 1 + \frac{1}{2} + 3 f(3) = 12 f(3) \]
Combining terms gives:
\[ \frac{3}{2} + 3 f(3) = 12 f(3) \]
Rearranging yields:
\[ \frac{3}{2} = 12 f(3) - 3 f(3) \implies \frac{3}{2} = 9 f(3) \implies f(3) = \frac{3}{2} \cdot \frac{1}{9} = \frac{1}{6} \]

### Step 4: Identify a pattern
From the calculations:
- \( f(1) = 1 \)
- \( f(2) = \frac{1}{4} \)
- \( f(3) = \frac{1}{6} \)

We can see a pattern emerging:
\[ f(n) = \frac{1}{2n} \]
This can be verified by checking the next values.

### Step 5: General formula for \( f(n) \)
Assuming \( f(n) = \frac{1}{2n} \) holds for all \( n \):
\[ f(n) = \frac{1}{2n} \]

### Step 6: Calculate \( f(1063) \)
Now we can find \( f(1063) \):
\[ f(1063) = \frac{1}{2 \cdot 1063} = \frac{1}{2126} \]

### Step 7: Calculate \( 2126 \cdot f(1063) \)
Now we compute:
\[ 2126 \cdot f(1063) = 2126 \cdot \frac{1}{2126} = 1 \]

### Final Answer
The value of \( 2126 f(1063) \) is \( \boxed{1} \).