To prove the equation $1 + 2 + 2^2 + 2^3 + \dots + 2^{n-1} = 2^n - 1$ using mathematical induction, we'll follow the standard steps of induction: the base case, the induction hypothesis, and the induction step.

Step 1: Base Case

First, let's verify the base case when $n = 1$.

For $n = 1$, the left-hand side of the equation is: $1$ The right-hand side of the equation is: $2^1 - 1 = 2 - 1 = 1$ Since both sides are equal, the base case holds.

Step 2: Induction Hypothesis

Assume that the formula holds for some positive integer $k$. That is, assume: $1 + 2 + 2^2 + 2^3 + \dots + 2^{k-1} = 2^k - 1$

Step 3: Induction Step

We need to show that the formula also holds for $k + 1$. That is, we need to prove: $1 + 2 + 2^2 + 2^3 + \dots + 2^{k-1} + 2^k = 2^{k+1} - 1$

Start by adding $2^k$ to both sides of the induction hypothesis: $1 + 2 + 2^2 + 2^3 + \dots + 2^{k-1} + 2^k = (2^k - 1) + 2^k$ Simplifying the right-hand side: $(2^k - 1) + 2^k = 2^k + 2^k - 1 = 2 \cdot 2^k - 1 = 2^{k+1} - 1$

This shows that if the formula holds for $n = k$, it also holds for $n = k + 1$.

Conclusion

By the principle of mathematical induction, the formula $1 + 2 + 2^2 + 2^3 + \dots + 2^{n-1} = 2^n - 1$ is true for all positive integers $n$.

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Would you like more details on any specific part of this proof, or do you have any other questions?

Here are 5 related questions to explore:

1. How does mathematical induction apply to inequalities?
2. Can you prove the sum of a geometric series using induction?
3. How does the inductive step change if we alter the base case?
4. What is the difference between strong induction and regular induction?
5. How would the proof change if the series started with a different initial term?

Tip:

When using mathematical induction, always clearly distinguish between the assumption (induction hypothesis) and what you need to prove (induction step). This helps in structuring the proof logically.