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.