Question

Prove by induction that $$ 1+r+r^{2}+\cdots+r^{k}+\cdots+r^{n}=\frac{1-r^{n+1}}{1-r} $$

Short answer

The formula holds true by induction.

Step-by-step solution

- Understand the Problem

The goal is to prove by induction that the sum of a geometric series from 1 to r^n is given by the formula: \[ 1 + r + r^2 + \text{...} + r^n = \frac{1 - r^{n+1}}{1 - r} \].

- Base Case

For the base case, let n = 0. The left-hand side of the equation is 1. Substituting n = 0 in the right-hand side gives: \[ \frac{1 - r^{0+1}}{1 - r} = \frac{1 - r}{1 - r} = 1 \]. Both sides are equal, so the base case holds.

- Inductive Hypothesis

Assume that the statement is true for some arbitrary positive integer k, i.e., assume \[ 1 + r + r^2 + \text{...} + r^k = \frac{1 - r^{k+1}}{1 - r} \].

- Inductive Step

We need to show the statement holds for k + 1. Consider the sum up to k + 1: \[ 1 + r + r^2 + \text{...} + r^k + r^{k+1} \]. Using the inductive hypothesis, rewrite the sum: \[ \frac{1 - r^{k+1}}{1 - r} + r^{k+1} \] Factor out \( \frac{1}{1 - r} \) from both terms: \[ \frac{1 - r^{k+1} + r^{k+1}(1 - r)}{1 - r} \].

- Simplify

Simplify the numerator: \[ 1 - r^{k+1} + r^{k+1} - r^{k+2} \] which simplifies to \[ 1 - r^{k+2} \]. Thus, the expression becomes: \[ \frac{1 - r^{k+2}}{1 - r} \]. This matches the right-hand side of the formula for n = k + 1.

- Conclusion

By the principle of mathematical induction, the formula \[ 1 + r + r^2 + \text{...} + r^n = \frac{1 - r^{n+1}}{1 - r} \] holds for all non-negative integers n.

Key concepts

geometric series

A geometric series is a sequence of terms where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio, denoted as \( r \). Geometric series can be finite or infinite depending on the number of terms in the series.
For example, a geometric series with the common ratio \( r \) and starting with 1 can be written as: \[ 1, r, r^2, r^3, \text{...} \text{,} r^n \] where \( n \) is the number of terms - 1.
A great thing about geometric series is we can easily find the sum of its terms using the formula:
\[ 1 + r + r^2 + \text{...} + r^n = \frac{1 - r^{n+1}}{1 - r} \]
This sum formula is crucial for dealing with problems related to geometric series in algebra and calculus.

inductive hypothesis

The inductive hypothesis is a crucial step in mathematical induction. It's the assumption made that a statement is true for some arbitrary positive integer \( k \). This forms the basis around which you prove that the statement holds for \( k + 1 \).
To consider it in the context of our problem, we assume the formula for the sum of a geometric series holds for \( n = k \), i.e.
\[ 1 + r + r^2 + \text{...} + r^k = \frac{1 - r^{k+1}}{1 - r} \]
This assumption helps bridge the gap between proving the base case and proving the statement for all natural numbers incrementally.

base case

The base case is the initial step in the process of mathematical induction. It involves proving that the statement to be inducted is true for the initial value of the variable, usually \( n = 0 \) or \( n = 1 \).
In this problem, the base case begins with \( n = 0 \). The left-hand side of our geometric series formula is just the first term, which is 1.
Substituting \( n = 0 \) in the right-hand side of our formula, we get:
\[ \frac{1 - r^{0+1}}{1 - r} = \frac{1 - r}{1 - r} = 1 \]
Both sides of the equation are equal, thereby establishing our base case. This initial verification is the foundational support needed for the inductive step.

inductive step

The inductive step is where you use the inductive hypothesis to prove that the statement holds for \( n = k + 1 \). By showing this, we can conclude that the statement is true for all natural numbers following the base case.
Given our inductive hypothesis:
\[ 1 + r + r^2 + \text{...} + r^k = \frac{1 - r^{k+1}}{1 - r} \]
We need to show that the statement holds for \( n = k + 1 \). Consider the sum up to \( k + 1 \):
\[ 1 + r + r^2 + \text{...} + r^k + r^{k+1} \]
Using the inductive hypothesis, we can rewrite the sum:
\[ \frac{1 - r^{k+1}}{1 - r} + r^{k+1} \]
We then factor out the common factor \( \frac{1}{1 - r} \):
\[ \frac{1 - r^{k+1} + r^{k+1}(1 - r)}{1 - r} \]
Simplify the numerator:
\[ 1 - r^{k+1} + r^{k+1} - r^{k+2} \]
Which becomes:
\[ 1 - r^{k+2} \]
Thus, the expression simplifies to the formula:
\[ \frac{1 - r^{k+2}}{1 - r} \]
This matches the right-hand side of the formula for \( n = k + 1 \), completing the inductive step.