Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ 3+3^{2}+3^{3}+\cdots+3^{n}=\frac{3}{2}\left(3^{n}-1\right) $$

Step-by-Step Solution

Answer

The statement is true for every positive integer \(n\) by induction.

Step 1: Understand the Statement

The exercise asks us to prove the equality \(3+3^2+3^3+\cdots+3^n=\frac{3}{2}(3^n-1)\) for every positive integer \(n\). This is a summation of a geometric series formula that we need to prove using mathematical induction.

Step 2: Base Case Verification

For mathematical induction, we start by verifying the base case, \(n = 1\). Substitute \(n = 1\) into the equation: LHS becomes \(3\), and RHS becomes \(\frac{3}{2}(3^1-1) = \frac{3}{2}\times2 = 3\). Thus, both sides are equal, verifying the base case.

Step 3: Inductive Hypothesis

Assume the statement holds for some positive integer \(k\), i.e., \(3+3^2+3^3+\cdots+3^k = \frac{3}{2}(3^k-1)\). This assumption will help us prove the next step.

Step 4: Inductive Step

We need to prove that if the statement holds for \(k\), then it also holds for \(k+1\). Start from the left side:\(3+3^2+\cdots+3^k + 3^{k+1}\).Using the inductive hypothesis, rewrite it as:\(\frac{3}{2}(3^k-1) + 3^{k+1}\).

Step 5: Simplify the Inductive Step

Continue the simplification:\(= \frac{3}{2}(3^k-1) + 3 \cdot 3^k\)\(= \frac{3}{2} \cdot 3^k - \frac{3}{2} + 3^{k+1}\)\(= \frac{3 \cdot 3^k}{2} - \frac{3}{2} + 3 \cdot 3^k\)\(= \frac{3 \cdot 3^k}{2} + \frac{6 \cdot 3^k}{2} - \frac{3}{2} \)\(= \frac{9 \cdot 3^k}{2} - \frac{3}{2} \)\(= \frac{3(3^{k+1}-1)}{2}\). Thus, the statement holds for \(k+1\).

Step 6: Conclusion

Since the base case \(n=1\) is true and if true for \(k\) implies true for \(k+1\), by mathematical induction, the statement holds for all positive integers \(n\).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the given exercise, the sequence starts with 3 and each subsequent term is multiplied by 3, making 3 the common ratio:

• First term: 3
• Second term: 3²
• Third term: 3³
• And so on...

Geometric series have a specific formula for the sum of the first n terms, which helps us understand and compute the total without having to do every single addition manually. The formula for the sum of the first n terms of a geometric series (where the first term is \(a\) and the common ratio is \(r\)) is:\[S_n = a \frac{r^n - 1}{r - 1}\]In this case, our formula simplifies to the specific problem as \(3 + 3^2 + 3^3 + \cdots + 3^n = \frac{3}{2}(3^n - 1)\), which we aim to prove using induction.

Positive Integers

Positive integers are the whole numbers greater than zero and are used in many mathematical theorems and proofs. They are fundamental building blocks in mathematics and can be denoted as the set \(\mathbb{Z}^+\). These numbers appear frequently in sequences, series, and various proof techniques, such as the one in our exercise, where \(n\) is a positive integer.Working with positive integers often simplifies mathematical problems since they eliminate fractions or negatives and keep calculations straightforward. In the exercise, the proof requires showing the validity of the given formula for any positive integer \(n\), which sets a clear and simple starting point since the operation and properties are well-defined and understood.

Proof Techniques

Proof techniques in mathematics are methods used to demonstrate the truth of a statement. One common proof technique is mathematical induction, which is especially useful for statements about natural numbers or sequences. Here’s a breakdown of how mathematical induction typically works:- **Base Case**: Verify the statement for the initial value, often \(n=1\). This acts as the foundation.
In our example, substituting \(n=1\) yields equal sides of the equation and confirms that the base case holds. - **Inductive Hypothesis**: Assume the statement is true for some positive integer \(k\). This is a crucial step where we make a leap that if true for one \(n\), it should be true for \(k\).
In the solution, it assumes the formula holds for \(n=k\).- **Inductive Step**: Use the assumption to prove that if the statement is true for \(k\), it holds for \(k+1\). This connects the hypothesis to the broader statement, essentially claiming the statement's truth for all subsequent values.
Thus, simplifying the expression by showing that given \(k\), it seamlessly steps to \(k+1\) corrects any leaps in logic.Mathematical induction, when done correctly, provides a strong and valid argument for proving statements involving positive integers or series.