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The problem asks for a proof using mathematical induction to show that for all positive integers \(n\):

\[
2 + 4 + 6 + \cdots + 2n = n(n + 1)
\]

Let's break it down following the steps provided:

### Step 1: Base Case (Prove for \(n = 1\))

For \(n = 1\), the left-hand side (LHS) is just:
\[
2 = 2
\]
The right-hand side (RHS) is:
\[
1(1 + 1) = 1 \times 2 = 2
\]
Thus, the base case holds true: 
\[
LHS = RHS
\]

### Step 2: Inductive Hypothesis (Assume it's true for \(n = k\))

Assume that for some \(n = k\), the statement holds true. That is, assume:
\[
2 + 4 + 6 + \cdots + 2k = k(k + 1)
\]
This is the inductive hypothesis.

### Step 3: Inductive Step (Prove for \(n = k + 1\))

We need to prove that the statement holds for \(n = k + 1\). That is, we want to show:
\[
2 + 4 + 6 + \cdots + 2k + 2(k + 1) = (k + 1)((k + 1) + 1)
\]

Using the inductive hypothesis, we know that:
\[
2 + 4 + 6 + \cdots + 2k = k(k + 1)
\]
Now, add \(2(k + 1)\) to both sides:
\[
k(k + 1) + 2(k + 1) = (k + 1)(k + 2)
\]

Simplifying the right-hand side:
\[
(k + 1)(k + 2) = (k + 1)(k + 1 + 1)
\]
Thus, the inductive step holds, and by the principle of mathematical induction, the statement is true for all positive integers \(n\).

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Do you have any questions or need further details?

Here are 5 related questions to further your understanding:
1. What is the principle of mathematical induction and how does it work?
2. How would you apply induction to prove a different summation formula?
3. Why do we need both the base case and inductive step for mathematical induction?
4. What are some common mistakes to avoid when using mathematical induction?
5. How can you verify your inductive hypothesis in more complex problems?

**Tip:** When using induction, always check your inductive hypothesis carefully to ensure it's set up correctly!

Prove by mathematical induction that for all positive integers n, 2 + 4 + 6 + ... + 2n = n(n+1).

Learn how to use mathematical induction to prove that the summation of even numbers 2 + 4 + 6 + ... + 2n equals n(n+1). This step-by-step solution involves proving the base case, assuming for n=k, and proving for n=k+1. Suitable for high school students in Grades 10-12.

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