Math Problem Statement

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

Solution

To prove that n2<3nn^2 < 3^n for all n{0,1,2,3,}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 nn.

  • For n=0n = 0: n2=02=0,3n=30=1n^2 = 0^2 = 0, \quad 3^n = 3^0 = 1 Clearly, 0<10 < 1, so the inequality holds for n=0n = 0.

  • For n=1n = 1: n2=12=1,3n=31=3n^2 = 1^2 = 1, \quad 3^n = 3^1 = 3 Clearly, 1<31 < 3, so the inequality holds for n=1n = 1.

  • For n=2n = 2: n2=22=4,3n=32=9n^2 = 2^2 = 4, \quad 3^n = 3^2 = 9 Clearly, 4<94 < 9, so the inequality holds for n=2n = 2.

  • For n=3n = 3: n2=32=9,3n=33=27n^2 = 3^2 = 9, \quad 3^n = 3^3 = 27 Clearly, 9<279 < 27, so the inequality holds for n=3n = 3.

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

Step 2: Inductive step

We assume that the inequality holds for some arbitrary n=kn = k. That is, we assume: k2<3kk^2 < 3^k This is called the inductive hypothesis.

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

Expand both sides:

  • The left-hand side: (k+1)2=k2+2k+1(k + 1)^2 = k^2 + 2k + 1

  • The right-hand side: 3k+1=33k3^{k+1} = 3 \cdot 3^k

Thus, we need to prove that: k2+2k+1<33kk^2 + 2k + 1 < 3 \cdot 3^k

Using the inductive hypothesis:

Since we know that k2<3kk^2 < 3^k, we can replace k2k^2 with a smaller value: k2+2k+1<3k+2k+1k^2 + 2k + 1 < 3^k + 2k + 1 So, now we need to show that: 3k+2k+1<33k3^k + 2k + 1 < 3 \cdot 3^k

Simplifying the right-hand side: 33k=3k+23k3 \cdot 3^k = 3^k + 2 \cdot 3^k Thus, we need to show: 3k+2k+1<3k+23k3^k + 2k + 1 < 3^k + 2 \cdot 3^k

Now, cancel 3k3^k from both sides: 2k+1<23k2k + 1 < 2 \cdot 3^k

Since 3k3^k grows exponentially, it is easy to verify that 2k+1<23k2k + 1 < 2 \cdot 3^k for all k3k \geq 3. Therefore, the inequality holds for n=k+1n = k + 1.

Step 3: Conclusion

By the principle of mathematical induction, we have shown that n2<3nn^2 < 3^n for all n0n \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 2n>n32^n > n^3 for large nn?
  2. What other methods could be used to compare exponential and polynomial growth?
  3. How does n2n^2 compare with 4n4^n as nn increases?
  4. What are some practical applications of comparing polynomial and exponential growth?
  5. Can we establish bounds for the smallest nn where n23nn^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.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Exponential Growth
Polynomial Growth
Inequalities

Formulas

n^2 < 3^n
Inductive hypothesis: k^2 < 3^k
Expanding: (k + 1)^2 = k^2 + 2k + 1
Exponential term: 3^{k+1} = 3 * 3^k

Theorems

Principle of Mathematical Induction

Suitable Grade Level

Grades 11-12