Math Problem Statement

Use induction to show that 2^n ≥ n^2 + ∀ n ≥ 5

Solution

To prove 2nn22^n \geq n^2 for all n5n \geq 5 using mathematical induction, we need to follow three steps: the base case, the induction hypothesis, and the induction step.

Step 1: Base Case (n=5n = 5)

For n=5n = 5: 25=32and52=25.2^5 = 32 \quad \text{and} \quad 5^2 = 25. Clearly, 322532 \geq 25, so the base case holds.

Step 2: Induction Hypothesis

Assume that 2kk22^k \geq k^2 is true for some integer k5k \geq 5. That is, we assume: 2kk2.2^k \geq k^2.

Step 3: Induction Step

We need to show that 2k+1(k+1)22^{k+1} \geq (k+1)^2 given that 2kk22^k \geq k^2.

Start with the left-hand side of the inequality: 2k+1=22k.2^{k+1} = 2 \cdot 2^k.

By the induction hypothesis, we know 2kk22^k \geq k^2, so we substitute: 2k+12k2.2^{k+1} \geq 2 \cdot k^2.

Now, we need to show that: 2k2(k+1)2.2 \cdot k^2 \geq (k+1)^2.

Expanding (k+1)2(k+1)^2: (k+1)2=k2+2k+1.(k+1)^2 = k^2 + 2k + 1.

So, it is enough to show: 2k2k2+2k+1.2k^2 \geq k^2 + 2k + 1.

Simplify this inequality: k22k+1.k^2 \geq 2k + 1.

This can be rewritten as: k22k10.k^2 - 2k - 1 \geq 0.

Step 4: Verify for k5k \geq 5

Now, we need to check if the quadratic k22k1k^2 - 2k - 1 is non-negative for k5k \geq 5.

The quadratic k22k1k^2 - 2k - 1 can be solved using the quadratic formula: k=2±4+42=2±82=1±2.k = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}.

The roots are approximately 120.411 - \sqrt{2} \approx -0.41 and 1+22.411 + \sqrt{2} \approx 2.41. Since k5k \geq 5, this inequality holds true for all k5k \geq 5.

Thus, k22k10k^2 - 2k - 1 \geq 0 for k5k \geq 5, which means: 2k2(k+1)2for all k5.2k^2 \geq (k+1)^2 \quad \text{for all } k \geq 5.

Conclusion

By induction, we have shown that 2nn22^n \geq n^2 for all n5n \geq 5.

Would you like more details or have any questions?

Here are 5 related questions:

  1. How does the quadratic formula help solve inequalities in induction proofs?
  2. What are some other examples where induction is useful in proving inequalities?
  3. Can induction be used to prove geometric series inequalities?
  4. How can we verify induction results using graphing techniques?
  5. Why is the base case important in an induction proof?

Tip: When using induction, always ensure that the induction hypothesis is correctly assumed before moving to the next step.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Exponential Growth
Quadratic Equations
Inequalities

Formulas

Induction Hypothesis: 2^k ≥ k^2
Base Case: 2^n = 32 and n^2 = 25 for n = 5
Induction Step: 2^{k+1} = 2 * 2^k, (k+1)^2 = k^2 + 2k + 1
Quadratic Inequality: k^2 - 2k - 1 ≥ 0

Theorems

Mathematical Induction
Quadratic Formula

Suitable Grade Level

Grades 10-12