Let ` A = { n in N | n^(2) le n + 10,000 } B = { 3k + 1 | k in N } and C = { 2 k |k in K }` , then the sum of all the elements of the set ` A cap (B - C)` is equal to ______

Answer

To solve the problem, we need to find the sum of all elements in the set \( A \cap (B - C) \). Let's break this down step by step.

### Step 1: Define the Set A
The set \( A \) is defined as:
\[ A = \{ n \in \mathbb{N} \mid n^2 \leq n + 10,000 \} \]
Rearranging the inequality gives:
\[ n^2 - n - 10,000 \leq 0 \]
To find the roots of the quadratic equation \( n^2 - n - 10,000 = 0 \), we can use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -10,000 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-10,000) = 1 + 40,000 = 40,001 \]
Now, applying the quadratic formula:
\[ n = \frac{1 \pm \sqrt{40,001}}{2} \]
Calculating \( \sqrt{40,001} \) approximately gives \( 200 \) (since \( 200^2 = 40,000 \)). Thus:
\[ n \approx \frac{1 \pm 200}{2} \]
This gives two roots:
\[ n_1 \approx 100.5 \quad \text{and} \quad n_2 \approx -99.5 \]
Since \( n \) must be a natural number, we take \( n \leq 100 \). Therefore, the set \( A \) is:
\[ A = \{ 1, 2, 3, \ldots, 100 \} \]

### Step 2: Define the Set B
The set \( B \) is defined as:
\[ B = \{ 3k + 1 \mid k \in \mathbb{N} \} \]
Calculating the first few elements:
- For \( k = 1 \): \( 3(1) + 1 = 4 \)
- For \( k = 2 \): \( 3(2) + 1 = 7 \)
- For \( k = 3 \): \( 3(3) + 1 = 10 \)
- For \( k = 4 \): \( 3(4) + 1 = 13 \)
- Continuing this way, we get:
\[ B = \{ 4, 7, 10, 13, 16, \ldots \} \]

### Step 3: Define the Set C
The set \( C \) is defined as:
\[ C = \{ 2k \mid k \in \mathbb{N} \} \]
Calculating the first few elements:
- For \( k = 1 \): \( 2(1) = 2 \)
- For \( k = 2 \): \( 2(2) = 4 \)
- For \( k = 3 \): \( 2(3) = 6 \)
- Continuing this way, we get:
\[ C = \{ 2, 4, 6, 8, 10, 12, \ldots \} \]

### Step 4: Find \( B - C \)
To find \( B - C \), we remove the elements of \( C \) from \( B \):
\[ B - C = \{ 7, 13, 19, 25, \ldots \} \]
This set consists of all elements of \( B \) that are not even numbers. The general form of this set can be expressed as:
\[ B - C = \{ 6k + 1 \mid k \in \mathbb{N} \} \]

### Step 5: Find \( A \cap (B - C) \)
Now we need to find the intersection \( A \cap (B - C) \):
\[ A \cap (B - C) = \{ 7, 13, 19, 25, \ldots, 97 \} \]
These are the odd numbers in \( A \) that are also in \( B - C \) and less than or equal to 100.

### Step 6: Sum the Elements of \( A \cap (B - C) \)
The sequence \( 7, 13, 19, \ldots, 97 \) is an arithmetic progression (AP) where:
- First term \( a = 7 \)
- Common difference \( d = 6 \)

To find the number of terms \( n \):
\[ a_n = a + (n-1)d \implies 97 = 7 + (n-1)6 \]
Solving for \( n \):
\[ 97 - 7 = (n-1)6 \implies 90 = (n-1)6 \implies n - 1 = 15 \implies n = 16 \]

Now, we can find the sum \( S_n \) of the first \( n \) terms of an AP:
\[ S_n = \frac{n}{2} (a + l) \]
where \( l \) is the last term. Thus:
\[ S_{16} = \frac{16}{2} (7 + 97) = 8 \times 104 = 832 \]

### Final Answer
The sum of all elements of the set \( A \cap (B - C) \) is:
\[ \boxed{832} \]