To prove the statement

$1 + 2 + 2^2 + 2^3 + \dots + 2^n = 2^{n+1} - 1$

using mathematical induction, we proceed in two steps:

Step 1: Base Case

We start by verifying the formula for $n = 1$.

For $n = 1$: $1 + 2 = 2^{1+1} - 1 = 2^2 - 1 = 4 - 1 = 3$

Thus, the base case holds since $1 + 2 = 3$.

Step 2: Inductive Step

Assume that the statement is true for some $n = k$; that is, we assume

$1 + 2 + 2^2 + 2^3 + \dots + 2^k = 2^{k+1} - 1$

This is our inductive hypothesis. We need to show that this assumption implies the statement is true for $n = k+1$, i.e.,

$1 + 2 + 2^2 + 2^3 + \dots + 2^k + 2^{k+1} = 2^{(k+1)+1} - 1 = 2^{k+2} - 1$

Starting from the left-hand side, we add $2^{k+1}$ to both sides of the inductive hypothesis:

$1 + 2 + 2^2 + 2^3 + \dots + 2^k + 2^{k+1} = (2^{k+1} - 1) + 2^{k+1}$

Now, simplify the right-hand side:

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

Thus, we have shown that if the statement holds for $n = k$, it also holds for $n = k+1$.

Conclusion

By mathematical induction, the formula is true for all $n \geq 1$:

$1 + 2 + 2^2 + 2^3 + \dots + 2^n = 2^{n+1} - 1$

Would you like additional details on any part of the solution?

---

Related Questions

1. How does mathematical induction work for other series?
2. Can this formula be derived using a different approach?
3. How does induction apply to inequalities instead of equalities?
4. What other types of mathematical proofs complement induction?
5. What are some common applications of summing powers of 2?

Tip: In mathematical induction, always check that the base case holds; it ensures that the pattern starts correctly.