prove that n^2 < 3^n for every n e {0,1,2,3,...}

To prove that \( n^2 < 3^n \) for all \( n \in \{0, 1, 2, 3, \dots \} \), we can approach this problem using **mathematical induction**. Here's a step-by-step proof:

### Step 1: Base case
We first check if the inequality holds for small values of \( n \).

- For \( n = 0 \):
  \[
  n^2 = 0^2 = 0, \quad 3^n = 3^0 = 1
  \]
  Clearly, \( 0 < 1 \), so the inequality holds for \( n = 0 \).

- For \( n = 1 \):
  \[
  n^2 = 1^2 = 1, \quad 3^n = 3^1 = 3
  \]
  Clearly, \( 1 < 3 \), so the inequality holds for \( n = 1 \).

- For \( n = 2 \):
  \[
  n^2 = 2^2 = 4, \quad 3^n = 3^2 = 9
  \]
  Clearly, \( 4 < 9 \), so the inequality holds for \( n = 2 \).

- For \( n = 3 \):
  \[
  n^2 = 3^2 = 9, \quad 3^n = 3^3 = 27
  \]
  Clearly, \( 9 < 27 \), so the inequality holds for \( n = 3 \).

Thus, the base cases \( n = 0, 1, 2, 3 \) are true.

### Step 2: Inductive step

We assume that the inequality holds for some arbitrary \( n = k \). That is, we assume:
\[
k^2 < 3^k
\]
This is called the **inductive hypothesis**.

Now, we must prove that the inequality also holds for \( n = k + 1 \). In other words, we need to show:
\[
(k + 1)^2 < 3^{k+1}
\]

#### Expand both sides:

- The left-hand side:
  \[
  (k + 1)^2 = k^2 + 2k + 1
  \]

- The right-hand side:
  \[
  3^{k+1} = 3 \cdot 3^k
  \]

Thus, we need to prove that:
\[
k^2 + 2k + 1 < 3 \cdot 3^k
\]

#### Using the inductive hypothesis:
Since we know that \( k^2 < 3^k \), we can replace \( k^2 \) with a smaller value:
\[
k^2 + 2k + 1 < 3^k + 2k + 1
\]
So, now we need to show that:
\[
3^k + 2k + 1 < 3 \cdot 3^k
\]

Simplifying the right-hand side:
\[
3 \cdot 3^k = 3^k + 2 \cdot 3^k
\]
Thus, we need to show:
\[
3^k + 2k + 1 < 3^k + 2 \cdot 3^k
\]

Now, cancel \( 3^k \) from both sides:
\[
2k + 1 < 2 \cdot 3^k
\]

Since \( 3^k \) grows exponentially, it is easy to verify that \( 2k + 1 < 2 \cdot 3^k \) for all \( k \geq 3 \). Therefore, the inequality holds for \( n = k + 1 \).

### Step 3: Conclusion
By the principle of **mathematical induction**, we have shown that \( n^2 < 3^n \) for all \( n \geq 0 \).

---

Would you like more details on any step, or have any questions?

Here are some related questions to explore:

1. How can we prove that \( 2^n > n^3 \) for large \( n \)?
2. What other methods could be used to compare exponential and polynomial growth?
3. How does \( n^2 \) compare with \( 4^n \) as \( n \) increases?
4. What are some practical applications of comparing polynomial and exponential growth?
5. Can we establish bounds for the smallest \( n \) where \( n^2 \geq 3^n \)?

**Tip:** Induction is a powerful tool to prove statements about all natural numbers, especially when dealing with inequalities that involve sequences or recursive relationships.

Prove that n^2 < 3^n for every n ∈ {0, 1, 2, 3,...}

Explore how to prove the inequality n^2 < 3^n for all natural numbers n using the method of mathematical induction. This problem covers key mathematical concepts such as exponential and polynomial growth, and the principle of induction. Suitable for advanced high school students (Grades 11-12).

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