Consider a function f: IN`rarr`IR,satisfying `f(1)+2(f2)+3f(3)+ ... +xf(x) = x(x+1)f(x):x ge 2` with `f(1)=1`.Then `1/f(2022)+1/f(2028)` is equal to

Options

A. 8400
B. 8100
C. 8200
D. 8000

Answer

To solve the problem, we need to analyze the given function \( f: \mathbb{N} \to \mathbb{R} \) that satisfies the equation:

\[ f(1) + 2f(2) + 3f(3) + \ldots + xf(x) = x(x+1)f(x) \quad \text{for } x \geq 2 \]

with the condition \( f(1) = 1 \).

### Step 1: Establish the equation for \( x = n \)

Let’s denote the left-hand side of the equation as \( S_n = f(1) + 2f(2) + 3f(3) + \ldots + nf(n) \). Thus, we have:

\[ S_n = n(n+1)f(n) \]

### Step 2: Establish the equation for \( x = n + 1 \)

Now, let’s consider the case when \( x = n + 1 \):

\[ S_{n+1} = f(1) + 2f(2) + 3f(3) + \ldots + nf(n) + (n+1)f(n+1) = (n+1)(n+2)f(n+1) \]

### Step 3: Relate \( S_{n+1} \) and \( S_n \)

We can express \( S_{n+1} \) in terms of \( S_n \):

\[ S_{n+1} = S_n + (n+1)f(n+1) \]

### Step 4: Set up the equation

From the two expressions for \( S_n \) and \( S_{n+1} \), we have:

\[ S_n + (n+1)f(n+1) = (n+1)(n+2)f(n+1) \]

Substituting \( S_n = n(n+1)f(n) \):

\[ n(n+1)f(n) + (n+1)f(n+1) = (n+1)(n+2)f(n+1) \]

### Step 5: Simplify the equation

We can factor out \( (n+1) \):

\[ (n+1)f(n) + f(n+1) = (n+2)f(n+1) \]

Rearranging gives us:

\[ (n+1)f(n) = (n+2)f(n+1) - f(n+1) \]
\[ (n+1)f(n) = n f(n+1) \]

### Step 6: Establish a recursive relationship

This leads us to the relationship:

\[ \frac{f(n)}{f(n+1)} = \frac{n}{n+1} \]

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

From this relationship, we can express \( f(n) \) in terms of \( f(1) \):

\[ f(n) = \frac{1}{n} f(1) = \frac{1}{n} \]

### Step 8: Verify the function

Since \( f(1) = 1 \), we have:

\[ f(n) = \frac{1}{n} \]

### Step 9: Calculate \( \frac{1}{f(2022)} + \frac{1}{f(2028)} \)

Now we need to find:

\[ \frac{1}{f(2022)} + \frac{1}{f(2028)} = 2022 + 2028 = 4050 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{4050} \]