Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+3+3^{2}+\cdots+3^{n-1}=\frac{1}{2}\left(3^{n}-1\right) $$

Short answer

The formula is proven by induction: it holds for n = 1 and if it holds for n = k it holds for n = k + 1.

Step-by-step solution

- Base Case

Verify the statement for the initial value, usually for n = 1:When n = 1,The left-hand side (LHS) is: \[1\]The right-hand side (RHS) is:\[\frac{1}{2}(3^1 - 1) = \frac{1}{2}(3 - 1) = \frac{1}{2} \times 2 = 1\]Since LHS = RHS, the base case holds true.

- Induction Hypothesis

Assume that the statement holds true for some natural number k. That is, assume:\[1 + 3 + 3^2 + \, \cdots \, + 3^{k-1} = \frac{1}{2}(3^k - 1)\]

- Inductive Step

Show that the statement holds true for k + 1. Using the induction hypothesis, we need to prove:\[1 + 3 + 3^2 + \, \cdots \, + 3^{k-1} + 3^k = \frac{1}{2}(3^{k+1} - 1)\]Start with the induction hypothesis and add \(3^k\) to both sides:\[\frac{1}{2}(3^k - 1) + 3^k = \frac{1}{2}(3^k - 1) + \frac{2 \times 3^k}{2} = \frac{1}{2}(3^k - 1 + 2 \times 3^k) = \frac{1}{2}(3^k - 1 + 2 \times 3^k)\]Combine like terms inside the argument:\[\frac{1}{2}(3^k + 3^k) = \frac{1}{2}(3^k (1 + 2)) = \frac{1}{2}(3^{k+1} - 1)\]Thus, the statement is true for n = k + 1.

- Conclusion

Since the statement is true for n = 1 (base case), and the truth of the statement for n = k implies its truth for n = k + 1 (inductive step), by the Principle of Mathematical Induction, the given statement holds for all natural numbers n.

Key concepts

mathematical induction

Mathematical induction is a powerful proof technique used in mathematics to establish the validity of a statement for all natural numbers. The process involves two primary steps: the base case and the inductive step. The base case verifies the statement for the smallest natural number, usually 1 (or sometimes 0). This step ensures that the statement is true at the very start.

Next comes the inductive step. Here, we assume that the statement is true for some arbitrary natural number, say k. This assumption is known as the induction hypothesis. We then show that if the statement holds for n = k, it must also hold for n = k + 1. Together, these steps create a domino effect, proving the statement true for all natural numbers.

proof by induction

Proof by induction is a detailed application of mathematical induction where we meticulously follow the steps to prove a statement. Let's break it down further:

- **Base Case:** We start with the smallest value (usually n = 1). If the statement holds true for this initial value, we move to the next step.
- **Induction Hypothesis:** We assume that the statement is true for an arbitrary natural number k. This step is crucial as it sets the ground for the next part of the proof.
- **Inductive Step:** Using the induction hypothesis, we prove that the statement is true for k + 1. This often involves algebraic manipulation to reveal that the assumption for k implies correctness for k + 1.

Apply these steps correctly and you'll have a robust proof that leverages the principle of induction.

series summation

A series summation involves adding a sequence of terms that follow a specific pattern. Consider the given series in the example, 1 + 3 + 3^2 + ... + 3^{n-1}. This is a geometric series where each term is a multiple of the previous term by a constant factor (in this case, 3). The general formula for the sum of the first n terms of a geometric series is:
\[S_n = a \frac{r^n - 1}{r - 1}\]
where S_n is the sum, a is the first term, and r is the common ratio.

In our case, the first term is 1, and the common ratio r is 3. So by applying the formula, one can find the sum up to the nth term.
Understanding this summation formula is key to proving statements involving sequences and series.

algebraic proof

An algebraic proof involves manipulating algebraic expressions to show that a statement is true. In the inductive step of our exercise, we assume the statement is true for n = k:
\[1 + 3 + 3^2 + ... + 3^{k-1} = \frac{1}{2}(3^k - 1)\]
We then need to prove it true for n = k + 1 by adding the next term 3^k to both sides:
\[1 + 3 + 3^2 + ... + 3^{k-1} + 3^k = \frac{1}{2}(3^k - 1) + 3^k = \frac{1}{2}(3^{k+1} - 1)\]
Through a series of algebraic steps, we simplify and combine like terms, demonstrating that indeed our assumption extends to n = k + 1. This step-by-step manipulation is the core of an algebraic proof, where each logical step builds on the previous one to reach a conclusion.