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 \( f(2010) \) based on the given functional equation and initial condition.

### Step-by-Step Solution:

1. **Understand the Given Information:**
We know that:
- \( f(1) = 1 \)
- For \( n \geq 2 \), the equation is:
\[ f(1) + 2f(2) + 3f(3) + \ldots + nf(n) = n(n+1)f(n) \]

2. **Substituting \( n = 2 \):**
Substitute \( n = 2 \) into the equation:
\[ f(1) + 2f(2) = 2(2+1)f(2) \]
This simplifies to:
\[ 1 + 2f(2) = 6f(2) \]
Rearranging gives:
\[ 1 = 6f(2) - 2f(2) \implies 1 = 4f(2) \implies f(2) = \frac{1}{4} \]

3. **Substituting \( n = 3 \):**
Now substitute \( n = 3 \):
\[ f(1) + 2f(2) + 3f(3) = 3(3+1)f(3) \]
Plugging in the known values:
\[ 1 + 2 \cdot \frac{1}{4} + 3f(3) = 12f(3) \]
Simplifying gives:
\[ 1 + \frac{1}{2} + 3f(3) = 12f(3) \]
This simplifies to:
\[ \frac{3}{2} + 3f(3) = 12f(3) \]
Rearranging gives:
\[ \frac{3}{2} = 12f(3) - 3f(3) \implies \frac{3}{2} = 9f(3) \implies f(3) = \frac{1}{6} \]

4. **Substituting \( n = 4 \):**
Next, substitute \( n = 4 \):
\[ f(1) + 2f(2) + 3f(3) + 4f(4) = 4(4+1)f(4) \]
Plugging in the known values:
\[ 1 + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{6} + 4f(4) = 20f(4) \]
Simplifying gives:
\[ 1 + \frac{1}{2} + \frac{1}{2} + 4f(4) = 20f(4) \]
Thus:
\[ 2 + 4f(4) = 20f(4) \]
Rearranging gives:
\[ 2 = 20f(4) - 4f(4) \implies 2 = 16f(4) \implies f(4) = \frac{1}{8} \]

5. **Finding a Pattern:**
Continuing this process, we can observe a pattern:
- \( f(1) = 1 \)
- \( f(2) = \frac{1}{4} \)
- \( f(3) = \frac{1}{6} \)
- \( f(4) = \frac{1}{8} \)

It appears that:
\[ f(n) = \frac{1}{2n} \]

6. **Finding \( f(2010) \):**
Using the pattern, we find:
\[ f(2010) = \frac{1}{2 \cdot 2010} = \frac{1}{4020} \]

7. **Final Calculation:**
We need to find \( \frac{1}{2010f(2010)} \):
\[ \frac{1}{2010f(2010)} = \frac{1}{2010 \cdot \frac{1}{4020}} = \frac{4020}{2010} = 2 \]

### Conclusion:
The final answer is:
\[ \frac{1}{2010f(2010)} = 2 \]