To prove the statement $1 + 2 + 4 + \cdots + 2^n = 2^{n+1} - 1$ using mathematical induction, we will follow these steps:

Step 1: Base Case

First, we need to verify the statement for the base case $n = 0$.

For $n = 0$, the left-hand side of the equation is just the first term in the series:

$1 = 2^0$

The right-hand side is:

$2^{0+1} - 1 = 2^1 - 1 = 2 - 1 = 1$

Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Next, assume that the statement is true for some arbitrary positive integer $k$. That is, assume:

$1 + 2 + 4 + \cdots + 2^k = 2^{k+1} - 1$

This assumption is called the inductive hypothesis.

Step 3: Inductive Step

Now, we need to prove that the statement is true for $n = k + 1$. That is, we need to show:

$1 + 2 + 4 + \cdots + 2^k + 2^{k+1} = 2^{(k+1) + 1} - 1$

Starting from the inductive hypothesis, add $2^{k+1}$ to both sides:

$1 + 2 + 4 + \cdots + 2^k + 2^{k+1} = \left(2^{k+1} - 1\right) + 2^{k+1}$

Simplify the right-hand side:

$1 + 2 + 4 + \cdots + 2^k + 2^{k+1} = 2^{k+1} - 1 + 2^{k+1} = 2 \cdot 2^{k+1} - 1 = 2^{k+2} - 1$

This is exactly the form we need for $n = k + 1$, which is:

$2^{(k+1) + 1} - 1 = 2^{k+2} - 1$

Step 4: Conclusion

Since the base case holds, and the inductive step has been proven, by mathematical induction, the statement

$1 + 2 + 4 + \cdots + 2^n = 2^{n+1} - 1$

is true for all non-negative integers $n$.

Would you like more details on any specific part, or do you have any questions?

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Here are some related questions to consider:

1. What are the other common examples where mathematical induction is used?
2. How can mathematical induction be applied to inequalities?
3. What is the difference between weak and strong induction?
4. How does the sum of a geometric series relate to this problem?
5. Can you extend this formula to other types of series?

Tip: When using mathematical induction, always ensure that the base case is clearly established; it is the foundation upon which the entire proof is built.