Math Problem Statement

Simplify the product: (1 + 1/2)(1 + 1/3)(1 + 1/4) ... (1 + 1/(n-1))

Solution

The image shows a mathematical expression and a multiple-choice question written in another language. Let's analyze the expression:

Expression:

(1+12)(1+13)(1+14)(1+1n1)\left(1 + \frac{1}{2}\right) \cdot \left(1 + \frac{1}{3}\right) \cdot \left(1 + \frac{1}{4}\right) \cdot \ldots \cdot \left(1 + \frac{1}{n-1}\right)

The goal is to simplify or find a pattern in the product of these terms.

Simplifying the Expression

Each term inside the parentheses is of the form: 1+1k=k+1k1 + \frac{1}{k} = \frac{k + 1}{k} where kk ranges from 2 to n1n-1.

Thus, the product can be rewritten as: k=2n1(1+1k)=k=2n1k+1k\prod_{k=2}^{n-1} \left(1 + \frac{1}{k}\right) = \prod_{k=2}^{n-1} \frac{k + 1}{k}

Observing the Pattern

Notice that this is a telescoping product. When expanded, many terms will cancel out:

324354nn1\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \ldots \cdot \frac{n}{n-1}

All intermediate terms cancel, leaving:

n2\frac{n}{2}

Conclusion

The answer is n2\frac{n}{2}, which matches the option labeled as "C)" in the image.


Would you like to see further details or ask additional questions? Here are some related questions you might find interesting:

  1. How does the telescoping product technique work in general?
  2. What are some other examples of telescoping series or products?
  3. How can this concept be applied to infinite products or sums?
  4. What are some real-life applications of such products in mathematics or physics?
  5. How can we verify the correctness of the result using mathematical induction?

Tip: Understanding telescoping series and products can significantly simplify complex mathematical problems.

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Math Problem Analysis

Mathematical Concepts

Algebra
Telescoping Products
Simplification

Formulas

1 + 1/k = (k+1)/k
Telescoping product formula

Theorems

Telescoping series

Suitable Grade Level

Grades 9-12