Question

Use mathematical induction to prove that \(\sum_{j=1}^{n}(2 j-1)=n^{2}\)

Short answer

Short Answer: Using mathematical induction, we proved that the given statement \(\sum_{j=1}^{n}(2j-1)=n^{2}\) holds for all positive integers n. We first showed that the base case was true when n=1. Then we assumed the statement to be true for n=k and proved it to be true for n=k+1. By the principle of mathematical induction, the statement is true for all positive integers n.

Step-by-step solution

Base Case (n=1)

Let's check the given expression for n=1: \(\sum_{j=1}^{1}(2j-1)\). The sum consists of only one term: \(2(1) - 1 = 1\). Now, let's check if it equals to \(1^2\): \(1^2 = 1\). Since both expressions are equal, the base case is true.

Inductive Step - Assume true for n=k

Assume that the statement holds for n = k, meaning that: \(\sum_{j=1}^{k}(2j-1) = k^2\)

Show that it's true for n=k+1

Now we want to show that the statement also holds for n = k + 1. To do this, we add the (k+1)-th term to both sides of our assumption: \(\sum_{j=1}^{k+1}(2j-1) = k^2 + (2(k+1) -1)\) Let's simplify the expression on the right side: \(k^2 + (2k + 2 -1) = k^2 + 2k +1\) Now, we can use our assumption: \(\sum_{j=1}^{k}(2j-1) + (2(k+1) -1) = k^2 + 2k +1\) Since our assumption says that \(\sum_{j=1}^{k}(2j-1) = k^2\), we have: \(k^2 + 2k + 1 = (k+1)^2\) This is true, since \((k+1)^2\) can be expanded to \(k^2 + 2k +1\). Thus, the statement holds for n=k+1.

Conclusion

Since the statement holds for the base case (n=1) and we proved that it holds for n=k+1 assuming it holds for n=k, by the principle of mathematical induction, the statement holds for all positive integers n: \(\sum_{j=1}^{n}(2j-1)=n^{2}\)

Key concepts

Proofs in Number Theory

Proofs play a critical role in number theory, as they are the means through which mathematicians ensure the validity of their claims about numbers. Unlike a mere observation or conjecture, a proof is a logical argument demonstrating that a proposition must be true. Every step in a proof must be justified by a rule of logic or a previously established theorem.

In number theory, proofs varied from direct proofs, where one shows that certain conditions yield a particular result, to more complex ones such as proof by contradiction, where one assumes the opposite of what needs to be proven and finds a contradiction. Among the different types of proofs, mathematical induction is particularly powerful, allowing us to prove statements about an infinite number of cases by confirming the base case and showing that if one case holds, the next one does too. This is useful, for instance, when demonstrating properties of integers, like the formula for the sum of the first n odd numbers which is a classic exercise in mathematical induction.

Sigma Notation

Sigma notation is a convenient and compact way to represent long sums, especially those with a pattern or a rule that defines the terms. The Greek capital letter sigma, \( \Sigma \), is used to indicate that a sum is being taken over a range of values. Here's the anatomy of sigma notation:

• The sigma symbol itself, indicating that a sum is to be taken.
• Below the sigma, an expression that gives the starting value of the index.
• Above the sigma, the stopping value of the index.
• To the right of the sigma, the expression to be summed, which often includes the index.

The index of the sum acts much like the iterator in a loop in computer programming, taking on a sequence of values in succession. Each term of the sum is computed by substituting the index into the expression. This concise notation allows for a clear understanding of the sum's scope and terms involved.

Inductive Step

The inductive step is where the heart of a proof by mathematical induction lies. After establishing the base case, which shows that a statement holds for the initial value—in our example, \( n = 1 \)—the inductive step demonstrates that if the statement holds for an arbitrary selected integer \( k \) (the induction hypothesis), it must also hold for the next integer, \( k + 1 \).

If this step can be successfully shown, it confirms that the statement must be true for every integer following the base case, due to the domino effect - if one falls, the next will follow. To ensure the effectiveness of this step, one must clearly express the assumption (the case for \( k \) is true) and then use logical and algebraic manipulations to show the case for \( k + 1 \) also holds true, linking the two cases in a chain of reasoning. Following this principle solidifies the argument and confirms that the proposition is true for all natural numbers greater than or equal to the base case.