Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+2+2^{2}+\cdots+2^{n-1}=2^{n}-1 $$

Short answer

Using induction, we proved the statement is true for all natural numbers n.

Step-by-step solution

- Base Case

Verify that the statement is true for the initial value, usually for n = 1. Substitute n = 1 into the formula: Left-Hand Side: \(1 = 2^{0} = 1\)Right-Hand Side: \(2^{1} - 1 = 2 - 1 = 1\)Since both sides are equal, the base case is true.

- Induction Hypothesis

Assume that the statement is true for some natural number k, i.e., \(1 + 2 + 2^2 + \cdots + 2^{k-1} = 2^k - 1\). This is the induction hypothesis.

- Induction Step

Prove that the statement is true for n = k + 1. Start from the induction hypothesis:By adding \(2^k\) to both sides of the hypothesis:\(1 + 2 + 2^2 + \cdots + 2^{k-1} + 2^k\)Using the hypothesis, the left-hand side becomes:\(2^k - 1 + 2^k\)Combine like terms:\(2^k - 1 + 2^k = 2 \cdot 2^k - 1 = 2^{k+1} - 1\)Thus, the statement is true for n = k + 1.

- Conclusion

Since the base case is true and the induction step has been proven, by the principle of mathematical induction, the statement \(1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1\) is true for all natural numbers n.

Key concepts

Proof by Induction

Mathematical induction is a powerful proof technique used to show the truth of an infinite number of cases, often involving natural numbers.
The process involves two main steps: the base case and the inductive step.
By proving the base case, you demonstrate that the statement holds for the initial value, usually for n = 1.
Then, in the inductive step, you assume the statement is true for some arbitrary natural number k and use this assumption to prove it for k+1.
This method leverages the principle that if a domino falls, it will knock over the next domino.
By ensuring the first domino (base case) falls and showing that if one domino falls, the next will also fall (inductive step), you can conclude that all the dominos (statements) will fall.

Base Case

The base case is the initial step in a proof by induction.
It shows that the statement you are trying to prove is true for the smallest value, usually n = 1.
In the given exercise, we substitute n = 1 into the formula and verify that both sides of the equation are equal.
For the equation, \(1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1\), we check:
- Left-Hand Side (LHS) with n = 1: \(1 = 2^0 = 1\)
- Right-Hand Side (RHS) with n = 1: \(2^1 - 1 = 2 - 1 = 1\).
Since LHS = RHS, the base case holds true.

Inductive Step

The inductive step is where you show that if the statement holds for an arbitrary natural number k, then it also holds for k+1.
This step relies on the induction hypothesis.
In the exercise, assume the statement is true for some natural number k: \(1 + 2 + 2^2 + \cdots + 2^{k-1} = 2^k - 1\).
Then, you need to prove it for k+1 by adding \(2^k\) to both sides of the hypothesis:
\(1 + 2 + 2^2 + \cdots + 2^{k-1} + 2^k \).
Using the hypothesis, the left-hand side becomes: \(2^k - 1 + 2^k \).
Combine the terms to get:
\(2^k - 1 + 2^k = 2 \cdot 2^k - 1 = 2^{k+1} - 1\).
Hence, you have shown the statement holds for k+1.

Natural Numbers

Natural numbers are the set of positive integers beginning from 1, 2, 3, and so on.
They are commonly used in proofs by induction because the technique leverages the inherent sequential nature of these numbers.
When proving statements involving sums, sequences, or properties indexed by natural numbers, proofs by induction are highly effective.
In our exercise, the goal was to prove that the sum of a sequence involving powers of 2 for any natural number n equals \(2^n - 1\).
By confirming this for the base case and proving it for every subsequent number using the inductive step, the proof covers all natural numbers.