Question

Use mathematical induction to prove that $$ \begin{aligned} a+(a+d)+(a+2 d) & \\ +\cdots+[a+(n-1) d] &=n a+d \frac{n(n-1)}{2} \end{aligned} $$

Short answer

Formula holds for all \( n \) by mathematical induction.

Step-by-step solution

Base Case

Verify that the statement holds for the smallest value of n, which is \( n = 1 \). Substitute \( n = 1 \) into both sides of the equation: Left-hand side: \( a \) Right-hand side: \( n a + d \frac{n(n-1)}{2} \rightarrow 1a + d \frac{1(1-1)}{2} = a \)Both sides are equal, so the base case holds.

Inductive Hypothesis

Assume that the statement holds for an arbitrary positive integer \( k \). That is, assume:\[ a + (a+d) + (a+2d) + \cdots + [a+(k-1)d] = k a + d \frac{k(k-1)}{2} \]

Inductive Step

Prove that the statement is true for \( n = k+1 \). Start with:\[ a + (a+d) + (a+2d) + \cdots + [a+kd] \]By the inductive hypothesis, this equals:\[ ka + d \frac{k(k-1)}{2} \]Add the next term \( a + kd \) to both sides:\[ a + (a+d) + (a+2d) + \cdots + [a+kd] = ka + d \frac{k(k-1)}{2} + (a + kd) \]Simplify the right-hand side:\[ ka + d \frac{k(k-1)}{2} + a + kd = (k+1)a + d \frac{k(k-1) + 2k}{2} \]\[ (k+1)a + d \frac{k^2 - k + 2k}{2} = (k+1)a + d \frac{k^2 + k}{2} \]\[ (k+1)a + d \frac{k(k+1)}{2} \]

Conclusion

We've shown that if the formula holds for \( n = k \), it must also hold for \( n = k+1 \). Since the base case is true, by mathematical induction, the formula holds for all positive integers \( n \).

Key concepts

algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. In the given problem, algebraic manipulation plays a crucial role. You often start with an equation and use algebra to transform and simplify it. For example, in step 3 of our solution, we add terms and combine like terms to simplify the equation. This involves the use of basic algebraic operations such as addition, multiplication, and factoring.

series and sequences

Series and sequences are fundamental concepts in mathematics that deal with ordered lists of numbers. A sequence is a list of numbers written in a specific order, and a series is the sum of the terms of a sequence.
In the given problem, we are dealing with an arithmetic series, which is a sequence where the difference between consecutive terms is constant. An example is the series given by: \(a, a+d, a+2d, ..., a+(n-1)d\). Understanding sequences and their sums (series) is essential in solving problems involving patterns and progressions.

proof techniques

Proof techniques are methods used to establish the truth of mathematical statements. Mathematical induction, which we use in the given problem, is one such technique. It involves proving a base case for the smallest value (usually 1) and then proving that if the statement holds for some integer \(k\), it must also hold for \(k+1\). Other common proof techniques include direct proof, proof by contradiction, and proof by contraposition. Each technique has its unique approach and utility in different scenarios.

base case

The base case is the initial step in a proof by induction. It verifies that the statement holds for the smallest possible value, often \(n = 1\). In our example, we check the base case by substituting \(n = 1\) into both sides of the equation. We find:

• Left-hand side: \(a\)
• Right-hand side: \(1a + d \frac{1(1-1)}{2} = a\)

Both sides are equal, proving that the base case holds true. This initial step is crucial because it provides the foundation for the induction process.

inductive hypothesis

The inductive hypothesis is an assumption made in the second step of mathematical induction. You assume that the statement is true for an arbitrary positive integer \(k\). In our example, we assume:
\(a + (a+d) + (a+2d) + ... + [a+(k-1)d] = ka + d \frac{k(k-1)}{2}\).
This assumption is the basis for proving the next step, where we show the statement holds for \(k+1\). The inductive hypothesis bridges the base case and the inductive step, making it a critical component of the induction process.

inductive step

The inductive step involves proving that if the statement holds for some integer \(k\), it must also hold for \(k+1\). In the given problem, we start with the series up to \(k+1\):
\(a + (a+d) + (a+2d) + ... + [a+kd]\).
By the inductive hypothesis, this equals:
\(ka + d \frac{k(k-1)}{2}\).
We then add the next term \(a + kd\) to both sides and simplify:
\(a + (a+d) + (a+2d) + ... + [a+kd] = ka + d \frac{k(k-1)}{2} + (a + kd)\).
This simplifies to:
\((k+1)a + d \frac{k(k+1)}{2}\).
Hence, we've shown that if the statement holds for \(k\), it also holds for \(k+1\). This completes the inductive step and the overall proof by mathematical induction.