Problem 44 - Algebra and Trigonometry

Question

Find \(S_{1}\) through \(S_{5}\) and then use the pattern to make a conjecture about \(S_{n}\). Prove the conjectured formula for \(S_{n}\) by mathematical induction. \(S_{n}:\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{n+1}\right)=?\)

Step-by-Step Solution

Answer

The conjectured formula for the series is \(S_{n} = \frac{1}{n+1}\). After calculating the first five terms, this formula was proven to be true using mathematical induction.

Step 1: Calculate the first five terms

Plug values from 1 to 5 into the given formula \( S_{n}:\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots \left(1-\frac{1}{n+1}\right) \) to obtain the first five terms.

Step 2: Make a conjecture for \(S_{n}\)

Observe the pattern of the first five terms and make a logical conjecture for \(S_{n}\). In this case, we notice that the series promotes the pattern of \(S_{n} = \frac{1}{n+1}\).

Step 3: Prove Base Case

To prove the conjecture by mathematical induction, we first show it is true for \(n=1\). Substituting \(n=1\) into both sides of the conjecture, it can be seen that they are equal, thus proving the base case.

Step 4: Assume True for n=k

Next, the inductive assumption is made where the conjecture is assumed to be true for \(n=k\). From this, we product the statement \(S_{k} =\frac{1}{k+1}\).

Step 5: Prove True for n=k+1

To prove the formula holds for \(n=k+1\), substitute \(n=k+1\) into the conjectured formula. Manipulate the result to show it is the same as the formula with \(n=k\). This proves the conjecture due to the principle of mathematical induction.

Key Concepts

Conjecture

In mathematics, forming a conjecture is the process of identifying a pattern or trend based on a limited set of examples or observations. For the exercise given, we calculated the first five terms of a sequence and observed a pattern. Conjecturing is like making an educated guess. We speculate on what the general rule for all terms might be. In this case, after examining the pattern of terms, we gathered the hypothesis that the sequence formula is \(S_{n} = \frac{1}{n+1}\). This conjecture needs to be proven true for all \(n\) using mathematical induction to ensure the validity across an entire series, not just the small initial set of terms observed.

Series and Sequences

Series and sequences are fundamental concepts in mathematics, particularly in understanding patterns and analyzing mathematical sets. A sequence is essentially a list of numbers in a specific order, following a certain rule. For example, the sequence given in this exercise consists of fractions decreasing by an increment in the denominator each time. The sequence's formula is \((1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})\ldots(1-\frac{1}{n+1})\). Each term is derived based on its position in the sequence (the nth term), and how it relates to the previously derived terms.
Studying such sequences often involves summing terms together, creating a series. We recognize the emerging pattern in the series which is crucial because it triggers the observation that leads to our conjecture. Sequences help us in making predictions about future values as they reveal consistent properties and behaviors of numbers.

Base Case

When proving a conjecture using mathematical induction, the base case is our starting point. We first demonstrate that the formula holds true for the smallest possible value of \(n\), typically \(n=1\). In this exercise, substituting \(n=1\) into the conjectured formula \(S_{n} = \frac{1}{n+1}\) results in \(S_{1} = \frac{1}{2}\). Similarly, calculating directly from the series formula, \((1-\frac{1}{2})\), also results in \(\frac{1}{2}\).
Confirming the base case establishes that our conjecture works for the initial instance, providing a foundation upon which the rest of the proof rests. Without a valid base case, the inductive process cannot proceed to validate the conjecture for all positive integers.

Inductive Step

The inductive step is crucial in mathematical induction as it extends the truth of the conjecture beyond the base case. Here, we assume our formula \(S_{n} = \frac{1}{n+1}\) is valid for an arbitrary positive integer \(k\). This is called the inductive hypothesis. From this assumption, we aim to demonstrate that the formula also holds true for \(n = k+1\).
In this exercise, by plugging \(k+1\) into our formula to obtain \(S_{k+1}\), we work through the algebraic manipulations necessary to prove that this step leads consistently to the expression \(\frac{1}{k+2}\). Successfully completing this step verifies that if the formula holds for one term, it must logically hold for the next term as well. Thus, through this rigorous method, we ensure the conjecture is true for all natural numbers.