Question

Let \(P_{n}\) be the proposition that $$ 1+2+\cdots+n=\frac{(n+2)(n-1)}{2} $$ (a) Show that \(P_{n}\) implies \(P_{n+1}\). (b) Is there an integer \(n\) for which \(P_{n}\) is true?

Short answer

Short Answer: We used the principle of mathematical induction to prove that if \(P_n\) is true, then it implies that \(P_{n+1}\) is also true. However, when we tried to find an integer n for which the proposition holds, we were unable to find any values that satisfy the equation. Therefore, there is no integer n for which \(P_n\) is true, and the given proposition is false for all integer values of n.

Step-by-step solution

Recall the principle of mathematical induction

Mathematical induction is a principle used to prove statements involving an integer variable, typically denoted by n. The process consists of two steps: (1) Proving the statement for a base case, typically when n equals the smallest value in the domain, and (2) assuming the statement is true for an arbitrary value of n and then proving that the statement is also true for n+1.

Prove that \(P_{n}\) implies \(P_{n+1}\)

Assume that \(P_n\) is true, meaning that the following equation holds:$$ 1+2+\cdots+n=\frac{(n+2)(n-1)}{2} $$Now, we want to prove \(P_{n+1}\), meaning that the following equation holds:$$ 1+2+\cdots+n+(n+1)=\frac{((n+1)+2)((n+1)-1)}{2} $$In order to prove this, we will add \((n+1)\) to both sides of the assumed true equation for \(P_n\):$$ 1+2+\cdots+n+(n+1)=\frac{(n+2)(n-1)}{2}+(n+1) $$ Now we have to manipulate the right side of the equation to match the right side of the \(P_{n+1}\) equation.

Simplify the equation from Step 2

Our current equation is:$$ 1+2+\cdots+n+(n+1)=\frac{(n+2)(n-1)}{2}+(n+1) $$To match the right side of the \(P_{n+1}\) equation, we will take a common denominator and simplify the expression:$$ 1+2+\cdots+n+(n+1)=\frac{(n+2)(n-1)+2(n+1)}{2} $$Simplifying the numerator, we get:$$ 1+2+\cdots+n+(n+1)=\frac{n^2 - n + 2n + 4}{2} $$Which simplifies further to:$$ 1+2+\cdots+n+(n+1)=\frac{(n+3)(n+1)}{2} $$Now the right side of our equation matches the right side of the \(P_{n+1}\) equation, so we have proven that \(P_n\) implies \(P_{n+1}\).

Find an integer n for which \(P_n\) is true

Now we need to find an integer \(n\) for which the given proposition holds. Let's try \(n = 1\), and check if the equation holds:$$ 1=\frac{(1+2)(1-1)}{2} $$This simplifies to:$$ 1=\frac{3 \cdot 0}{2}=0 $$Since 1 is not equal to 0, the proposition is not true for \(n=1\). Let's try \(n = 2\):$$ 1+2=\frac{(2+2)(2-1)}{2} $$This simplifies to:$$ 3=\frac{4 \cdot 1}{2}=2 $$Again, the proposition is not true for \(n=2\). As we keep trying different integers for n, we find that there is no integer n for which \(P_n\) holds. Therefore, the answer to part (b) is that no, there is no integer n for which \(P_n\) is true. The given proposition is false for all integer values of n.

Key concepts

proposition

In mathematics, a **proposition** denotes a statement or assertion that can be proved or disproved. It generally encapsulates an idea or claim based on mathematical logic. The proposition in our exercise is \[ P_{n}: 1 + 2 + \cdots + n = \frac{(n+2)(n-1)}{2} \] This proposition suggests an equality that involves the sum of a sequence of integers from 1 to \( n \) and a formula involving \( n \). The proposition is neither trivially true nor false. Instead, it requires proof or disproof, using logical reasoning or mathematical techniques. When a proposition involves an integer variable like \( n \), proving it often involves mathematical induction.
A crucial part of understanding any proposition is to evaluate if it is universally true, true under certain conditions, or false altogether. In our exercise, we investigate the truth of the proposition for various integer values of \( n \). Unfortunately, in this case, the proposition turns out to be false, as it does not hold for any tested integer values. This aspect highlights the importance of carefully checking the validity of mathematical propositions across defined domains.

integer variable

An **integer variable** in mathematics is a symbol that represents an unknown element within the set of integers, which are whole numbers that can be positive, negative, or zero. In the context of mathematical induction, the variable \( n \) commonly acts as the integer variable. It indicates a specific position within a sequence or a specific instance of a mathematical statement to be proved.
In our exercise, \( n \) represents, for example, the highest number in the series 1 to \( n \). Each time \( n \) takes on a different integer value, it presents a unique proposition. This flexibility allows us to explore how the equality holds or fails for each integer.
When proving propositions by mathematical induction, the integer variable is crucial. It helps us to verify the base case, and extend that judgment to all subsequent integers in the sequence, as shown by the step assuming \( P_n \) implies \( P_{n+1} \). This step in induction allows us to see if properties of a smaller whole number extend logically to larger numbers, based on our proof.

base case

The **base case** is a fundamental concept in mathematical induction. It is the starting point or smallest possible value for the integer variable, where the proposition is verified directly, without assumption. Proving the base case lays the groundwork to proceed with proving the proposition for larger values of the integer.
In the given exercise, the base case would typically be defined as \( n = 1 \). Checking whether \( n = 1 \) results in a true statement, however, reveals an inconsistency in the equality:\[ 1 = \frac{(1+2)(1-1)}{2} = 0 \] Therefore, the proposition fails at \( n = 1 \), showing it is not true, which causes the entire proof to crumble at its first step.
The failure of the base case means that no further conclusions can be drawn about higher numbers through induction because there's no initial proof to base subsequent steps on. Usually, resolving such issues requires revisiting assumptions or reformulating the proposition to find any potential oversight.