Problem 40 - Calculus: Early Transcendental Functions

Question

Use induction to derive the geometric series formula \(a+a r+a r^{2}+\cdots+a r^{n}=\frac{a-a r^{n+1}}{1-r}\) for constants \(a\) and \(r \neq 1.\)

Step-by-Step Solution

Answer

The geometric series formula \(a+ar+ar^{2}+...+ar^{n} = \frac{a-ar^{n+1}}{1-r}\ for constants a and r≠1 can be proven correct via mathematical induction by first proving it true for the base case \(n=0\), then assuming it is true for an arbitrary \(k\), and finally proving it true for \(k+1\). This completes the principle of mathematical induction, thus showing the equation is correct for all positive integers \(n\).

Step 1: Base Case

The first step in mathematical induction is always to prove the base case. Here, the base case is \(n=0\). We start with the left-hand side of our equation, which for \(n=0\) becomes \(a\). Now we plug \(n=0\) into the right-hand side of the equation, which simplifies to \(\frac{a-a*r^{0+1}}{1-r}=\frac{a-ar}{1-r}\). Simplifying this shows that both sides equal \(a\) when \(n=0\), thus the base case holds true.

Step 2: Induction Step - Assume true for \(k\)

In the induction step, we must assume the formula is valid for a certain \(k\) number. So, we have \(a+ar+ar^{2}+...+ ar^{k} = \frac{a-ar^{k+1}}{1-r}\)

Step 3: Induction Step - Prove true for \(k+1\)

Then, we need to show that the formula holds for \(k+1\). We consider the sum of the series up to \(k+1\) terms on the left-hand side and add the \((k+1)^{th}\) term, which results in: \(a+ar+ar^{2}+...+ ar^{k}+ar^{k+1}\). If the formula is true for \(k+1\), the right-hand side when we substitute \(k+1\) for \(n\) must equal to the left-hand side. We get: \(\frac{a-ar^{(k+1)+1}}{1-r}= \frac{a-ar^{k+2}}{1-r}\). When we substitute the assumption from step 2 into the left-hand side, we get \(\frac{a-ar^{k+1}}{1-r}+ ar^{k+1} = \frac{a-ar^{k+2}}{1-r}\), which simplifies to the right-hand side, hence proving that the statement is true for \(k+1\) if it's true for \(k\).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This series can be represented as \( a + ar + ar^2 + \dots + ar^n \), where \( a \) is the first term and \( r \) is the common ratio.
Geometric series are powerful tools in mathematics since they allow us to sum a potentially infinite number of terms succinctly. The sum of the first \( n+1 \) terms of a geometric series can be expressed as:

• If \( r eq 1 \):
  1. \( S = \frac{a - ar^{n+1}}{1-r} \)
• If \( r = 1 \):
  1. \( S = a(n+1) \)

Understanding this concept is vital for applying mathematical induction successfully to derive the above formula.

Induction Base Case

The induction base case is the first step in proving a statement using mathematical induction. Here, we must demonstrate that the formula holds for the initial term, often where \( n = 0 \).
For our geometric series formula, when \( n = 0 \), the series simplifies to just \( a \). The right-hand side of our equation \( \frac{a - ar^{0+1}}{1-r} \) must also simplify to \( a \).
By confirming that both the left-hand and right-hand sides equal \( a \) when \( n = 0 \), we establish that our base case is valid. This provides a solid foundation for proceeding with the inductive step.

Induction Step

The induction step involves two critical processes: assuming the formula holds for a certain \( k \), and then showing it also holds for \( k+1 \). This is often referred to as the assumption and induction hypothesis.
First, assume the formula is true for \( n = k \), which means:
\( a + ar + ar^2 + \ldots + ar^k = \frac{a - ar^{k+1}}{1-r} \).
This is our induction hypothesis.
Next, prove it for \( n = k+1 \) by adding \( ar^{k+1} \) to the sum assumed to be true for \( k \):
\( a + ar + ar^2 + \ldots + ar^k + ar^{k+1} \).
The goal is to show:\
\( \frac{a - ar^{k+2}}{1-r} = \frac{a - ar^{k+1}}{1-r} + ar^{k+1} \).
This often involves simplifying the right-hand side to demonstrate it equals the formula we assumed for \( k+1 \). Achieving this completes the induction step.

Mathematical Proof

A mathematical proof is a logical argument that demonstrates the truth of a mathematical statement. In the context of mathematical induction, the proof involves two main parts: proving the base case and the induction step.
For the geometric series formula, the mathematical proof utilizes induction to ensure the formula works for all natural numbers \( n \). Starting with the base case for \( n = 0 \), we confirm that the initial state satisfies the equation. Next, through the induction step, we establish that if the formula holds for a particular \( n = k \), it must also hold for \( n = k+1 \).
This two-step process ensures the statement is valid for all integers, giving us a complete and rigorous proof of the geometric series formula using mathematical induction.