Problem 26 - Algebra 2

Question

ARTHMETIC SERIES Use mathematical induction to prove the formula \(a_{1}+\left(a_{1}+d\right)+\left(a_{1}+2 d\right)+\cdots+\left[a_{1}+(n-1) d\right]=\frac{n}{2}\left[2 a_{1}+(n-1) d\right]\) for the sum of an arithmetic series.

Step-by-Step Solution

Answer

The formula is proven true for all positive integers using mathematical induction.

Step 1: Base Case

Start by verifying the formula for the smallest possible value of \( n \), which is 1. In this case, the arithmetic series consists of one term, \( a_1 \). So, the left-hand side is simply \( a_1 \). Substituting \( n=1 \) into the right-hand side gives \( \frac{1}{2} [2a_1 + (1-1)d] = a_1 \). Since both sides equal \( a_1 \), the base case holds true.

Step 2: Inductive Hypothesis

Assume that the formula holds for some arbitrary positive integer \( k \). This means we assume that the sum of the first \( k \) terms of the arithmetic series is: \[ S_k = a_1 + (a_1 + d) + \cdots + [a_1 + (k-1)d] = \frac{k}{2}[2a_1 + (k-1)d]. \] This is known as our inductive hypothesis.

Step 3: Inductive Step

Next, we need to prove that if the formula works for \( n = k \), it also works for \( n = k+1 \). The sum of the first \( k+1 \) terms can be written as the sum of the first \( k \) terms plus the \((k+1)\)th term: \[ S_{k+1} = S_k + [a_1 + kd]. \] By the inductive hypothesis, \( S_k = \frac{k}{2}[2a_1 + (k-1)d] \). Substitute this into the equation: \[ S_{k+1} = \frac{k}{2}[2a_1 + (k-1)d] + [a_1 + kd]. \] Simplify the equation to: \[ S_{k+1} = \frac{k[2a_1 + (k-1)d] + 2[a_1 + kd]}{2}. \] This simplifies further to: \[ S_{k+1} = \frac{k+1}{2}[2a_1 + kd]. \] The right-hand side of this matches the original formula with \( n = k+1 \), completing the inductive step.

Step 4: Conclusion

Since the formula holds for \( n = 1 \) (the base case) and assuming it holds for \( n = k \) leads to it holding for \( n = k+1 \) (inductive step), by mathematical induction, the formula is true for all positive integers \( n \).

Key Concepts

Arithmetic Series

An arithmetic series is a sequence of numbers where each term after the first is formed by adding a constant difference, known as the common difference, to the previous term. These types of sequences are linear and are expressed as follows:

• The first term is denoted as \(a_1\).
• The common difference is denoted as \(d\).
• The \(n\)th term is expressed as \(a_1 + (n-1)d\).

To find the sum of the first \(n\) terms of such a series, there is a specific formula: \[a_1 + (a_1 + d) + (a_1 + 2d) + \ldots + (a_1 + (n-1)d) = \frac{n}{2}[2a_1 + (n-1)d].\]Understanding this formula is critical as it helps to calculate sums quickly without adding each term. The sum consists of \(n\) terms, where the average of each term is calculated based on the arithmetic progression properties.

Inductive Hypothesis

The inductive hypothesis is an essential part of mathematical induction. This hypothesis is our "assumption step." In the context of this exercise, it involves assuming that a given formula is true for an arbitrary case \(n = k\). For example, we assume:

• The sum of the first \(k\) terms of an arithmetic series accurately matches the formula:
• \[S_k = \frac{k}{2}[2a_1 + (k-1)d].\]

By assuming this to be true, we set the stage to prove the formula for the next step, \(k+1\). This step is vital because it allows us to use previously established logic to transition from one case to the next in the induction process.

Inductive Step

The inductive step is where we extend the truth of our formula from the case \(n = k\) to \(n = k+1\). In this phase, we begin by expressing the sum of the first \(k+1\) terms:

• We start with the known sum \(S_k\) and add the \((k+1)\)th term: \[S_{k+1} = S_k + (a_1 + kd).\]
• Substitute the inductive hypothesis: \[S_k = \frac{k}{2}[2a_1 + (k-1)d].\]

After substituting, simplify the result to match the original formula but with \(n = k+1\). This shows how the arithmetic continues seamlessly. Simplifying further, we confirm:
\[S_{k+1} = \frac{k+1}{2}[2a_1 + kd].\]
This step bridges the assumption with the new case, ensuring the formula holds as we add another term.

Base Case

In mathematical induction, the base case is where we show the formula works for the very first integer in question, \(n = 1\) in this instance. This case acts as the foundation or the anchor for the entire proof. Here:

• Our arithmetic series consists solely of the first term, \(a_1\).
• When \(n = 1\), both sides of the formula result in \(a_1\):
  \(\text{LHS} = a_1\)
• \(\text{RHS} = \frac{1}{2}[2a_1 + (1-1)d] = a_1\)

Since both sides are equal, this confirms that the formula is initially valid. Establishing the base case ensures the chain reaction for subsequent truths, paving the way for the inductive step to apply.