Simplifying The Product (1 - 1/2^2)(1 - 1/3^2)...(1 - 1/100^2) A Step-by-Step Guide

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In this article, we will delve into the simplification of a fascinating mathematical expression. Our focus is on understanding the elegant pattern hidden within the product of terms in the form of (1 - 1/n^2), where 'n' ranges from 2 to 100. This type of problem often appears in mathematical competitions and highlights the importance of recognizing and utilizing algebraic identities. The given expression is:

(1 - 1/2^2)(1 - 1/3^2)(1 - 1/4^2) ... (1 - 1/100^2)

Our goal is to simplify this expression and arrive at a concise numerical answer. This involves a combination of algebraic manipulation and pattern recognition. Let’s embark on this mathematical journey together, breaking down each step to ensure a clear and comprehensive understanding. The beauty of mathematics lies in its ability to transform complex problems into simple, elegant solutions through the application of fundamental principles. By the end of this exploration, you will not only have the solution to this specific problem but also a broader understanding of how to approach similar mathematical challenges.

Understanding the Core Concept: Difference of Squares

The key to simplifying the expression lies in recognizing the difference of squares pattern. This fundamental algebraic identity states that a^2 - b^2 can be factored into (a - b)(a + b). Applying this to our expression, we can rewrite each term (1 - 1/n^2) as a difference of squares. Specifically, we see 1 as 1^2 and 1/n^2 as (1/n)^2. Therefore, we can rewrite each term as:

1 - 1/n^2 = 1^2 - (1/n)^2

Now, applying the difference of squares factorization, we get:

1^2 - (1/n)^2 = (1 - 1/n)(1 + 1/n)

This transformation is crucial because it breaks down each term into two simpler factors. By applying this to every term in the original product, we set the stage for significant simplification through cancellation. The elegance of this approach lies in its ability to convert a seemingly complex product into a series of manageable factors, paving the way for a straightforward solution. This technique is widely applicable in various mathematical contexts, making it a valuable tool in any problem-solver's arsenal. Mastering the difference of squares and its applications is essential for anyone looking to excel in mathematics.

Applying the Difference of Squares to the Expression

Now that we understand the difference of squares identity, let's apply it to our original expression. We replace each term (1 - 1/n^2) with its factored form (1 - 1/n)(1 + 1/n). This gives us:

(1 - 1/2^2)(1 - 1/3^2)(1 - 1/4^2) ... (1 - 1/100^2) = [(1 - 1/2)(1 + 1/2)][(1 - 1/3)(1 + 1/3)][(1 - 1/4)(1 + 1/4)] ... [(1 - 1/100)(1 + 1/100)]

This expansion transforms the initial product into a series of products, each involving two factors. Next, we simplify each factor within the brackets. For a term like (1 - 1/n), we find a common denominator, which is 'n', resulting in (n - 1)/n. Similarly, for a term like (1 + 1/n), we get (n + 1)/n. Applying this simplification to each factor, we rewrite the expression as:

[(2 - 1)/2 * (2 + 1)/2] * [(3 - 1)/3 * (3 + 1)/3] * [(4 - 1)/4 * (4 + 1)/4] * ... * [(100 - 1)/100 * (100 + 1)/100]

Further simplification yields:

[1/2 * 3/2] * [2/3 * 4/3] * [3/4 * 5/4] * ... * [99/100 * 101/100]

This step is critical as it sets the stage for the next phase of simplification, where we'll observe a beautiful pattern of cancellation. The transformation highlights the power of algebraic manipulation in revealing hidden structures within complex expressions. By breaking down each term and applying basic arithmetic, we've made the expression more manageable and ready for the next stage of simplification.

Observing the Cancellation Pattern

The expression we've arrived at, [1/2 * 3/2] * [2/3 * 4/3] * [3/4 * 5/4] * ... * [99/100 * 101/100], exhibits a fascinating pattern of cancellation. When we expand this product, we can rewrite it as:

(1/2) * (3/2) * (2/3) * (4/3) * (3/4) * (5/4) * ... * (99/100) * (101/100)

Notice that many terms in the numerators and denominators will cancel each other out. For instance, the '3' in the numerator of the first term (3/2) cancels with the '3' in the denominator of the second term (2/3). Similarly, the '2' in the numerator of the second term cancels with the '2' in the denominator of the first term. This pattern continues throughout the expression. To visualize this cancellation more clearly, let’s rearrange the terms slightly:

(1/2) * (2/3) * (3/4) * ... * (99/100)   *   (3/2) * (4/3) * (5/4) * ... * (101/100)

In the first group, * (1/2) * (2/3) * (3/4) * ... * (99/100)*, the numerator of each fraction cancels with the denominator of the preceding fraction. This leaves us with only the first denominator (2) and the last numerator (99). In the second group, (3/2) * (4/3) * (5/4) * ... * (101/100), a similar cancellation occurs, leaving us with the first numerator (3) and the last denominator (100). This remarkable cancellation significantly simplifies the expression, paving the way for a straightforward final calculation.

Performing the Cancellation and Final Calculation

After observing the cancellation pattern, we can now perform the cancellations. In the first part of the expression, (1/2) * (2/3) * (3/4) * ... * (99/100), all the intermediate terms cancel out, leaving us with 1/100. This is because the numerator of each fraction cancels with the denominator of the previous fraction, resulting in a telescoping product. Similarly, in the second part of the expression, (3/2) * (4/3) * (5/4) * ... * (101/100), the same cancellation pattern occurs, leaving us with 101/2. So, the expression simplifies to:

(1/100) * (101/2)

Now, we simply multiply these two fractions together:

(1/100) * (101/2) = 101/200

Therefore, the simplified form of the original expression is 101/200. This final calculation demonstrates the power of recognizing patterns and applying algebraic principles to simplify complex expressions. The journey from the initial product of differences of squares to the final fraction highlights the beauty and efficiency of mathematical techniques. By breaking down the problem into manageable steps and utilizing the difference of squares identity, we've arrived at a concise and elegant solution.

Therefore, the final answer is:

101/200

In summary, we successfully simplified the expression [(1 - 1/2^2)(1 - 1/3^2)(1 - 1/4^2) ... (1 - 1/100^2)] to 101/200. This was achieved by recognizing and applying the difference of squares identity, which allowed us to factor each term in the product. We then observed a significant cancellation pattern, where intermediate terms in the numerators and denominators canceled each other out. This cancellation led to a much simpler expression, which we easily calculated to arrive at the final answer.

The key takeaways from this exercise include:

  1. The Difference of Squares: Understanding and applying the difference of squares identity (a^2 - b^2 = (a - b)(a + b)) is a powerful tool in simplifying algebraic expressions.
  2. Pattern Recognition: Identifying patterns, such as the cancellation pattern in this problem, is crucial for efficient problem-solving.
  3. Step-by-Step Simplification: Breaking down a complex problem into smaller, manageable steps makes the solution process clearer and less daunting.

This problem exemplifies how a combination of algebraic manipulation and keen observation can lead to elegant solutions in mathematics. The techniques used here are applicable to a wide range of problems, making this a valuable exercise in mathematical thinking and problem-solving. By mastering these concepts, you can confidently tackle similar challenges and appreciate the beauty and power of mathematics.