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We are tasked with proving the following statement using mathematical induction:

\[
\sum_{i=1}^{n} 2^{i-1} = 2^n - 1 \quad \text{for all} \quad n \geq 1
\]

### Step 1: Base Case (n = 1)
For \(n = 1\), the left-hand side of the equation is:

\[
\sum_{i=1}^{1} 2^{i-1} = 2^{1-1} = 2^0 = 1
\]

The right-hand side is:

\[
2^1 - 1 = 2 - 1 = 1
\]

Since both sides are equal, the base case holds true.

### Step 2: Inductive Hypothesis
Assume that the formula holds for some arbitrary \(n = k\). That is, assume:

\[
\sum_{i=1}^{k} 2^{i-1} = 2^k - 1
\]

### Step 3: Inductive Step
We need to show that the formula holds for \(n = k + 1\), i.e., we need to prove that:

\[
\sum_{i=1}^{k+1} 2^{i-1} = 2^{k+1} - 1
\]

Start with the left-hand side for \(n = k + 1\):

\[
\sum_{i=1}^{k+1} 2^{i-1} = \sum_{i=1}^{k} 2^{i-1} + 2^{(k+1)-1} = \sum_{i=1}^{k} 2^{i-1} + 2^k
\]

Using the inductive hypothesis, substitute \( \sum_{i=1}^{k} 2^{i-1} = 2^k - 1 \):

\[
= (2^k - 1) + 2^k = 2^k + 2^k - 1 = 2 \cdot 2^k - 1 = 2^{k+1} - 1
\]

Thus, the formula holds for \(n = k+1\).

### Step 4: Conclusion
By the principle of mathematical induction, the formula holds for all \(n \geq 1\).

Do you have any questions or would you like more details on a particular part?

Here are 5 related questions for further practice:

1. Prove by induction that \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \) for all \( n \geq 1 \).
2. Use induction to show that \( 3^n - 1 \) is divisible by 2 for all \( n \geq 1 \).
3. Prove by induction that \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \).
4. Show by induction that \( 2^n > n^2 \) for \( n \geq 5 \).
5. Prove by induction that the Fibonacci sequence satisfies \( F_{n+2} = F_{n+1} + F_n \).

**Tip:** When performing induction proofs, always ensure your inductive step clearly transitions from the assumption to the \( n+1 \) case.

Prove with mathematical induction that for all n ≥ 1, Σ (i=1 to n) 2^(i-1) = 2^n - 1.

Learn how to prove the formula Σ (i=1 to n) 2^(i-1) = 2^n - 1 using mathematical induction. This step-by-step guide covers key concepts in series, exponents, and the principle of induction. Suitable for high school students in Grades 10-12.

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