Let `f: N to R` be a function satisfying the condition `f(1) + f(2)`……………. `+f(n) = n^(2) f(n) AA n in N`
If `f(2025) = 1/(2026)`, then f(1) = ………….

Answer

To solve the problem, we need to find the value of \( f(1) \) given that \( f: \mathbb{N} \to \mathbb{R} \) satisfies the condition:

\[ f(1) + f(2) + \ldots + f(n) = n^2 f(n) \quad \text{for all } n \in \mathbb{N} \]

We are also given that \( f(2025) = \frac{1}{2026} \).

### Step 1: Analyze the given condition for small values of \( n \)

1. **For \( n = 1 \)**:
\[ f(1) = 1^2 f(1) \implies f(1) = f(1) \quad \text{(This is trivially true)} \]

2. **For \( n = 2 \)**:
\[ f(1) + f(2) = 2^2 f(2) \implies f(1) + f(2) = 4f(2) \]
Rearranging gives:
\[ f(1) = 4f(2) - f(2) = 3f(2) \implies f(1) = 3f(2) \]

### Step 2: Continue with \( n = 3 \)

3. **For \( n = 3 \)**:
\[ f(1) + f(2) + f(3) = 3^2 f(3) \implies f(1) + f(2) + f(3) = 9f(3) \]
Substitute \( f(1) = 3f(2) \):
\[ 3f(2) + f(2) + f(3) = 9f(3) \implies 4f(2) + f(3) = 9f(3) \]
Rearranging gives:
\[ 4f(2) = 8f(3) \implies f(2) = 2f(3) \]

### Step 3: Continue with \( n = 4 \)

4. **For \( n = 4 \)**:
\[ f(1) + f(2) + f(3) + f(4) = 4^2 f(4) \implies f(1) + f(2) + f(3) + f(4) = 16f(4) \]
Substitute \( f(1) = 3f(2) \) and \( f(2) = 2f(3) \):
\[ 3(2f(3)) + 2f(3) + f(4) = 16f(4) \implies 6f(3) + 2f(3) + f(4) = 16f(4) \]
This simplifies to:
\[ 8f(3) + f(4) = 16f(4) \implies 8f(3) = 15f(4) \implies f(4) = \frac{8}{15}f(3) \]

### Step 4: Identify the pattern

From the calculations, we can see a pattern emerging:
- \( f(1) = 3f(2) \)
- \( f(2) = 2f(3) \)
- \( f(3) = \frac{f(1)}{6} \)
- \( f(4) = \frac{8}{15}f(3) \)

### Step 5: Use the given value \( f(2025) = \frac{1}{2026} \)

Assuming the pattern holds, we can express \( f(n) \) in terms of \( f(1) \):
- \( f(2) = \frac{f(1)}{3} \)
- \( f(3) = \frac{f(1)}{6} \)
- Continuing this pattern leads to \( f(n) = \frac{f(1)}{T_n} \) where \( T_n \) is the \( n \)-th triangular number.

### Step 6: Calculate \( f(2025) \)

The \( 2025 \)-th triangular number \( T_{2025} = \frac{2025 \times 2026}{2} \).

Thus:
\[ f(2025) = \frac{f(1)}{T_{2025}} = \frac{f(1)}{\frac{2025 \times 2026}{2}} = \frac{2f(1)}{2025 \times 2026} \]

Setting this equal to the given value:
\[ \frac{2f(1)}{2025 \times 2026} = \frac{1}{2026} \]

### Step 7: Solve for \( f(1) \)

Cross-multiplying gives:
\[ 2f(1) = 2025 \implies f(1) = \frac{2025}{2} \]

### Final Answer
\[ f(1) = 1012.5 \]