To show that \( n^2 + 100n + 4 \in O(n^3) \), we need to prove that there exist constants \( c > 0 \) and \( n_0 \geq 0 \) such that for all \( n \geq n_0 \), the inequality 

\[
n^2 + 100n + 4 \leq c \cdot n^3
\]

holds true. This is the formal definition of Big-O notation, which indicates that the function \( n^2 + 100n + 4 \) grows asymptotically no faster than \( n^3 \).

### Step-by-Step Proof

1. **Consider the expression:**

   \[
   n^2 + 100n + 4
   \]

2. **Find an upper bound for \( n^2 + 100n + 4 \) in terms of \( n^3 \):**

   Notice that \( n^2, 100n, \) and \( 4 \) are all smaller than \( n^3 \) for sufficiently large \( n \). Thus, we can express the sum of these terms in terms of \( n^3 \):

   \[
   n^2 \leq n^3 \quad \text{for } n \geq 1
   \]
   \[
   100n \leq 100n^2 \leq 100n^3 \quad \text{for } n \geq 1
   \]
   \[
   4 \leq 4n^3 \quad \text{for } n \geq 1
   \]

3. **Combine these inequalities:**

   \[
   n^2 + 100n + 4 \leq n^3 + 100n^3 + 4n^3 = (1 + 100 + 4)n^3 = 105n^3
   \]

4. **Choose constants \( c \) and \( n_0 \):**

   Here, we can choose \( c = 105 \) and \( n_0 = 1 \).

5. **Verify the inequality:**

   For all \( n \geq 1 \):

   \[
   n^2 + 100n + 4 \leq 105n^3
   \]

Thus, we have shown that for \( c = 105 \) and \( n_0 = 1 \), the function \( n^2 + 100n + 4 \leq c \cdot n^3 \) for all \( n \geq n_0 \).

### Conclusion

Since we found constants \( c = 105 \) and \( n_0 = 1 \) such that the inequality \( n^2 + 100n + 4 \leq 105n^3 \) holds for all \( n \geq 1 \), we have demonstrated that \( n^2 + 100n + 4 \in O(n^3) \).

Would you like more details on this proof or have any questions?

Here are 5 related questions that expand upon this information:

1. How would you prove that \( n^2 + 100n + 4 \in O(n^2) \)?
2. Can we find a smaller constant \( c \) that works for \( n^2 + 100n + 4 \in O(n^3) \)?
3. What is the difference between Big-O, Big-Theta, and Big-Omega notations?
4. How can we prove \( n^2 + 100n + 4 \notin O(n) \)?
5. How would the proof change if we needed to show \( n^2 + 100n + 4 \in O(n^2 \log n) \)?

**Tip:** When proving Big-O notation, always look for the dominant term in the expression, as it often determines the simplest upper bound.

Discover how to prove that n^2 + 100n + 4 ∈ Ο(n^3) using Big-O notation. Follow a step-by-step mathematical proof demonstrating that for sufficiently large n, n^2 + 100n + 4 grows no faster than n^3. Suitable for advanced undergraduate students and anyone interested in computational complexity.

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