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The exercise in the image involves proving formulas using **mathematical induction**. Let's go over one of these proofs.

### Problem 1: 
Prove using induction:  
\[ 2 + 4 + 6 + \cdots + 2n = n(n+1) \]

**Step 1: Base Case (n=1)**  
When \( n = 1 \):  
The left-hand side (LHS) is \( 2 \), and the right-hand side (RHS) is:  
\[ 1(1 + 1) = 1 \times 2 = 2 \]  
So, LHS = RHS for \( n = 1 \). The base case holds.

**Step 2: Inductive Hypothesis**  
Assume the formula is true for some \( n = k \), i.e.,  
\[ 2 + 4 + 6 + \cdots + 2k = k(k + 1) \]

**Step 3: Inductive Step (n = k+1)**  
We need to show the formula holds for \( n = k+1 \). That is:  
\[ 2 + 4 + 6 + \cdots + 2k + 2(k + 1) = (k + 1)((k + 1) + 1) \]

Starting with the left-hand side, add \( 2(k + 1) \) to both sides of the inductive hypothesis:
\[ k(k + 1) + 2(k + 1) \]
Factor out \( (k + 1) \):
\[ (k + 1)(k + 2) \]

This is exactly the right-hand side of the equation for \( n = k+1 \). Hence, the formula is proven true for \( n = k+1 \).

By the principle of mathematical induction, the formula holds for all \( n \).

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Would you like to go through the next problem in the list or have more details about this proof? Here are some additional questions:

1. How does mathematical induction work in general?
2. Can you show the proof for problem 2?
3. How would the proof differ if the series involved squares or cubes of numbers?
4. Is there a geometric way to interpret these sums?
5. What are the common pitfalls when using mathematical induction?

**Tip:** Always clearly state the inductive hypothesis, as it is the key assumption that bridges the base case and the inductive step.

Establish the formulas given below by mathematical induction.

This page provides step-by-step solutions to a set of mathematical induction problems involving arithmetic series, factorial sums, and exponential growth. Learn how to establish these formulas through mathematical induction, a powerful proof technique. Suitable for high school or early college students.

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