Math Problem Statement

Choose one of the following states and prove it using mathematical induction.Once you begin your actual proof, be sure to include the basis step, inductive hypothesis, and inductive step.  The inductive step should clearly show the inductive hypothesis and conclusion

Choose one of the following states and prove it using mathematical induction. Once you begin your actual proof, be sure to include the basis step, inductive hypothesis, and inductive step. The inductive step should clearly show the inductive hypothesis and conclusion.

Solution

The statement provided in the image is:

2+4+6++2n=n(n+1)2 + 4 + 6 + \dots + 2n = n(n + 1)

You are asking for a proof using mathematical induction, which involves three key steps: the basis step, the inductive hypothesis, and the inductive step. Let me break down the proof process:

1. Basis Step:

We need to verify that the statement holds true for n=1n = 1.

For n=1n = 1:

  • The left-hand side (LHS) of the equation is simply 22, since the sum contains only the first term 22.
  • The right-hand side (RHS) is 1(1+1)=1×2=21(1 + 1) = 1 \times 2 = 2.

Since LHS=RHS\text{LHS} = \text{RHS}, the statement is true for n=1n = 1.

2. Inductive Hypothesis:

Assume that the statement holds true for some n=kn = k, i.e.,

2+4+6++2k=k(k+1)2 + 4 + 6 + \dots + 2k = k(k + 1)

This is our inductive hypothesis.

3. Inductive Step:

We need to prove that if the statement holds for n=kn = k, then it also holds for n=k+1n = k + 1. In other words, we need to show:

2+4+6++2k+2(k+1)=(k+1)((k+1)+1)2 + 4 + 6 + \dots + 2k + 2(k+1) = (k+1)((k+1) + 1)

Using the inductive hypothesis, we know:

2+4+6++2k=k(k+1)2 + 4 + 6 + \dots + 2k = k(k + 1)

Now, add 2(k+1)2(k+1) to both sides:

2+4+6++2k+2(k+1)=k(k+1)+2(k+1)2 + 4 + 6 + \dots + 2k + 2(k+1) = k(k + 1) + 2(k + 1)

Factor the right-hand side:

k(k+1)+2(k+1)=(k+1)(k+2)k(k + 1) + 2(k + 1) = (k + 1)(k + 2)

Thus, we have:

2+4+6++2k+2(k+1)=(k+1)(k+2)2 + 4 + 6 + \dots + 2k + 2(k+1) = (k + 1)(k + 2)

This proves that the statement holds for n=k+1n = k + 1.

Conclusion:

By the principle of mathematical induction, since the statement is true for n=1n = 1 (basis step), and assuming it holds for n=kn = k implies that it holds for n=k+1n = k + 1 (inductive step), the statement is true for all n1n \geq 1.


Would you like any further clarifications or details about any of these steps?

Here are 5 related questions to deepen your understanding:

  1. Can you explain why the basis step is necessary in mathematical induction?
  2. How would the proof change if the sum involved odd numbers instead of even?
  3. What are some common mistakes when applying the inductive step?
  4. Can you apply mathematical induction to prove statements in geometry or other fields?
  5. How can you extend this proof to find the sum of a more general arithmetic progression?

Tip: When performing the inductive step, always clearly state the hypothesis and the additional term being added to the sum to avoid mistakes.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Summation of Even Numbers

Formulas

2 + 4 + 6 + ... + 2n = n(n + 1)

Theorems

Mathematical Induction Principle

Suitable Grade Level

Grades 10-12