To prove the statement $1 + 2 + 2^2 + \dots + 2^{n-1} = 2^n - 1$ by mathematical induction, we proceed as follows:

Step 1: Base Case

We check the statement for $n = 1$.

• Left-hand side (LHS): For $n = 1$, the sum is $1$ (since the series only has the first term, $2^{1-1} = 2^0 = 1$).
• Right-hand side (RHS): $2^1 - 1 = 2 - 1 = 1$.

Since LHS = RHS, the base case holds true.

Step 2: Inductive Hypothesis

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

$1 + 2 + 2^2 + \dots + 2^{k-1} = 2^k - 1$

Step 3: Inductive Step

We need to show that the statement is true for $k+1$, i.e., we want to prove:

$1 + 2 + 2^2 + \dots + 2^{k-1} + 2^k = 2^{k+1} - 1$

Start with the left-hand side of the equation for $n = k + 1$:

$1 + 2 + 2^2 + \dots + 2^{k-1} + 2^k$

By the inductive hypothesis, we know:

$1 + 2 + 2^2 + \dots + 2^{k-1} = 2^k - 1$

So, we can substitute this into the equation:

$2^k - 1 + 2^k$

Simplify the expression:

$2^k - 1 + 2^k = 2 \times 2^k - 1 = 2^{k+1} - 1$

This is exactly the right-hand side of the equation for $n = k + 1$. Thus, the inductive step holds.

Conclusion:

Since the base case holds, and the inductive step has been proven, by the principle of mathematical induction, the statement $1 + 2 + 2^2 + \dots + 2^{n-1} = 2^n - 1$ is true for all positive integers $n$.

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Would you like more details on any part of this proof, or do you have any questions? Here are some related questions you might find interesting:

1. How does mathematical induction differ from direct proof?
2. Can this approach be used to prove inequalities?
3. How does this formula relate to the sum of a geometric series?
4. Can you prove similar formulas using mathematical induction?
5. What are the limitations of using mathematical induction?

Tip: Always verify the base case carefully, as the entire inductive proof depends on its correctness.