Let `f:N rarr R` be such that f(1)=1 and f(1)+2f(2)+3f(3)+…+nf(n)=n(n+1)f(n), for `n ge 2, " then " `/(2010f(2010))` is ……….. .

Answer

To solve the problem, we need to find the value of \( \frac{1}{2010 f(2010)} \) given the conditions of the function \( f \).

### Step 1: Understand the given conditions
We know:
1. \( f(1) = 1 \)
2. The equation:
\[ f(1) + 2f(2) + 3f(3) + \ldots + nf(n) = n(n+1)f(n) \quad \text{for } n \geq 2 \]

### Step 2: Substitute \( n = 2 \)
Substituting \( n = 2 \) into the equation:
\[ f(1) + 2f(2) = 2(2+1)f(2) \]
This simplifies to:
\[ 1 + 2f(2) = 6f(2) \]
Rearranging gives:
\[ 1 = 6f(2) - 2f(2) = 4f(2) \]
Thus, we find:
\[ f(2) = \frac{1}{4} \]

### Step 3: Substitute \( n = 3 \)
Next, substitute \( n = 3 \):
\[ f(1) + 2f(2) + 3f(3) = 3(3+1)f(3) \]
This simplifies to:
\[ 1 + 2 \cdot \frac{1}{4} + 3f(3) = 12f(3) \]
Calculating \( 1 + \frac{1}{2} = \frac{3}{2} \), we have:
\[ \frac{3}{2} + 3f(3) = 12f(3) \]
Rearranging gives:
\[ \frac{3}{2} = 12f(3) - 3f(3) = 9f(3) \]
Thus, we find:
\[ f(3) = \frac{1}{6} \]

### Step 4: Identify the pattern
From the values we have calculated:
- \( f(1) = 1 = \frac{1}{1} \)
- \( f(2) = \frac{1}{4} = \frac{1}{2 \cdot 2} \)
- \( f(3) = \frac{1}{6} = \frac{1}{2 \cdot 3} \)

We can see a pattern forming:
\[ f(n) = \frac{1}{2n} \quad \text{for } n \geq 1 \]

### Step 5: Calculate \( f(2010) \)
Using the established pattern:
\[ f(2010) = \frac{1}{2 \cdot 2010} = \frac{1}{4020} \]

### Step 6: Find \( 2010 f(2010) \)
Now we calculate:
\[ 2010 f(2010) = 2010 \cdot \frac{1}{4020} = \frac{2010}{4020} = \frac{1}{2} \]

### Step 7: Find \( \frac{1}{2010 f(2010)} \)
Finally, we find:
\[ \frac{1}{2010 f(2010)} = \frac{1}{\frac{1}{2}} = 2 \]

### Final Answer
Thus, the value of \( \frac{1}{2010 f(2010)} \) is \( \boxed{2} \).