Question

Use mathematical induction to derive the following results: $$ \sum_{k=0}^{n} r^{k}=\frac{r^{n+1}-1}{r-1}, \quad \sum_{k=0}^{n} k=\frac{n(n+1)}{2} $$

Short answer

Both \(\sum_{k=0}^{n} r^{k}=\frac{r^{n+1}-1}{r-1}\) and \(\sum_{k=0}^{n} k=\frac{n(n+1)}{2}\) can be proved using mathematical induction.

Step-by-step solution

Proving First Equation

First, prove the equation \(\sum_{k=0}^{n} r^{k}=\frac{r^{n+1}-1}{r-1}\) using mathematical induction.

Base Case

Start with the base case when \( n = 0 \). The left side of the equation becomes \( r^0 \) which is 1. The right side becomes \( \frac{r^{0+1}-1}{r-1} = 1 \). Hence, the equation holds for \( n = 0 \).

Inductive Step for First Equation

Now, assume the equation holds for some positive integer \( n \). This means \(\sum_{k=0}^{n} r^{k} = \frac{r^{n+1}-1}{r-1}\). Next, you must prove that the equation holds for \( n + 1 \). Start by adding \( r^{n+1} \) to both sides of the equation: \(\sum_{k=0}^{n+1} r^{k} = \frac{r^{n+1}-1}{r-1}+r^{n+1} = \frac{r^{n+2}-1}{r-1}\). Hence, the equation holds for \( n+1 \).

Proving Second Equation

Next, prove the equation \(\sum_{k=0}^{n} k=\frac{n(n+1)}{2}\) using mathematical induction.

Base Case for Second Equation

Start with the base case when \( n = 0 \). The left side becomes 0. The right side becomes \( \frac{0(0+1)}{2} = 0 \). Hence, the equation holds for \( n = 0 \).

Inductive Step for Second Equation

Assume the equation holds for some positive integer \( n \). This means \(\sum_{k=0}^{n} k = \frac{n(n+1)}{2}\). You must prove it holds for \( n + 1 \). Add \( n + 1 \) to both sides and simplify: \(\sum_{k=0}^{n+1} k = \frac{n(n+1)}{2} + (n + 1) = \frac{(n+1)((n+1)+1)}{2}\). Hence, the equation holds for \( n + 1 \).

Key concepts

Geometric Series

A geometric series is a sum of terms in which each term is a constant multiple of the previous one. Consider a series like \(a, ar, ar^2, ar^3, \ldots\). Each term is obtained by multiplying the previous term by \(r\), the common ratio. A key formula related to geometric series is:

• If you sum from the 0th to the nth term, the formula is \( \sum_{k=0}^{n} r^{k} = \frac{r^{n+1} - 1}{r - 1} \) when \( r eq 1 \).

This formula is crucial because it simplifies the calculation of the sum of a geometric series, making computations more straightforward. In the proven formula, the numerator \((r^{n+1} - 1)\) represents the difference between the power term due to multiplication by \(r\) and 1. The denominator \((r - 1)\) serves to normalize this difference, aligning it with standard scaling terms.

Sum of Integers

The sum of a sequence of integers from 1 through \(n\) is straightforward, yet fundamental in mathematics. It is expressed with the formula \( \sum_{k=0}^{n} k = \frac{n(n+1)}{2} \). This allows understanding the total of consecutive numbers easily. Let's break it down:

• The term \(n(n+1)\) represents the multiplication cycle required for pairing, typical in arithmetic series.
• Dividing by 2 adjusts for the fact that each sum pairs two numbers equally spaced from the ends.

An example is when calculating the sum of numbers from 0 to 5, \(0 + 1 + 2 + 3 + 4 + 5\). Using the formula gives \( \frac{5(5+1)}{2} = 15 \), showing pairing efficiently adds integers.

Proof Techniques

Proof techniques are logical steps used to establish the truth of mathematical statements. Inductive proof is a powerful method used to prove statements about integers. It's ideal for proving claims that something holds true for all natural numbers. The basic structure involves two main parts:

• Base Case: You start by proving the theorem for the initial value, usually \(n=0\) or \(n=1\).
• Inductive Step: You assume the statement is true for some arbitrary integer \(n\), and then show it must also be true for \(n+1\).

Through these steps, mathematical proofs become easier to understand and logical arguments are systematically constructed, ensuring that results hold for every integer in series.

Inductive Reasoning

Inductive reasoning is a fundamental proof strategy used in mathematics to validate statements that apply across infinite numbers of cases. It seeks evidence through logical progression, focusing on specific cases to prove larger truths. The process is systematic, ensuring

• Each step is logically derived from the preceding one.
• Every apparent pattern or case observed can extend to others universally through assumed truth.

In the context of proving series formulas, inductive reasoning allows moving from a known base case to infer the behavior of subsequent cases. This enables mathematicians to conclusively demonstrate general formulas like those of geometric series and sums of integers by verifying the continuation pattern holds universally.