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The problem in the image asks to use **mathematical induction** to prove the following sum:

\[
1 + 2 + 3 + \cdots + n = 2^{n+1} - 1
\]

for all positive integers \( n \).

### Step-by-Step Induction Proof:

#### **Step 1: Base Case ( \( n = 1 \) )**

For \( n = 1 \):

\[
1 = 2^{1+1} - 1 = 2^2 - 1 = 4 - 1 = 3
\]

The base case does not seem to hold for this form. Let's recheck if the formula might be miswritten and correct it to the standard summation formula, which should be:

\[
1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}
\]

This is the correct formula for the sum of the first \( n \) positive integers. If you're still following the question exactly as stated, let me know, and we can fix the misalignment.

Would you like more details on this or another mathematical proof using the correct formula?

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### 5 Related Questions:
1. How do we prove the sum of the first \( n \) positive integers using induction?
2. What are the key steps to verifying a formula with mathematical induction?
3. Can we use another method besides induction to verify sums like this one?
4. How does the base case in induction help establish the validity of a proof?
5. What other types of problems can be solved using induction?

### Tip: 
When solving with induction, always check if the base case holds before proceeding with the inductive step.

Use mathematical induction to show that 1 + 2 + 3 + ... + n = 2^(n+1) - 1 for all positive integers n.

Explore how to use mathematical induction to prove the formula for the sum of the first n positive integers: 1 + 2 + 3 + ... + n = 2^(n+1) - 1. A step-by-step breakdown for high school students learning induction.

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