Problem 102 - Algebra and Trigonometry

Question

Challenge Problem If the terms of a sequence have the property that \(\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\cdots=\frac{a_{n-1}}{a_{n}},\) show that \(\frac{a_{1}^{n}}{a_{2}^{n}}=\frac{a_{1}}{a_{n+1}}\)

Step-by-Step Solution

Answer

The identity \(\frac{a_{1}^{n}}{a_{2}^{n}} = \frac{a_{1}}{a_{n+1}}\) holds.

Step 1: Identify the common ratio

Given the property \(\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{3}}{a_{4}} \dots = \frac{a_{n-1}}{a_{n}},\) let \(k = \frac{a_{1}}{a_{2}}.\)

Step 2: Express all terms using common ratio

Using the common ratio \(k,\) express each term in the sequence in terms of \(a_{1}\): \(a_{2} = \frac{a_{1}}{k},\) \(a_{3} = \frac{a_{2}}{k} = \frac{a_{1}}{k^2},\) and so on. Therefore, \(a_{n} = \frac{a_{1}}{k^{n-1}}.\)

Step 3: Express \(a_{n+1}\)

Similarly, for \(a_{n+1},\) we have \(a_{n+1} = \frac{a_{n}}{k} = \frac{a_{1}}{k^{n}}.\)

Step 4: Rewrite \(\frac{a_{1}^{n}}{a_{2}^{n}}\)

Using \(a_{2} = \frac{a_{1}}{k},\) the expression \(\frac{a_{1}^{n}}{a_{2}^{n}}\) can be rewritten as \(\frac{a_{1}^{n}}{\big( \frac{a_{1}}{k} \big)^{n}} = \frac{a_{1}^{n}}{\frac{a_{1}^{n}}{k^{n}}} = k^{n} = \frac{a_{1}}{a_{n+1}}.\)

Step 5: Final verification

Confirm that both sides of the identity match: \(\frac{a_{1}^{n}}{a_{2}^{n}} = \frac{a_{1}}{a_{n+1}}.\)

Key Concepts

Common Ratio

The common ratio is a key concept in geometric sequences. It represents the constant factor between consecutive terms in the sequence. For example, if you have a sequence where each term is half of the previous term, the common ratio is \( \frac{1}{2} \).
In the given exercise, we identify the common ratio as \( k = \frac{a_1}{a_2} \). This implies that each term in the sequence can be expressed relative to the previous term using this ratio.
Understanding the common ratio helps simplify and manipulate the terms in a sequence for further calculations. Knowing this, you can easily express any term in the sequence if the first term and the common ratio are known.

Sequence Terms

In geometric sequences, each term is a product of the previous term and the common ratio. For this problem, the sequence terms can be explicitly written by expressing each following term using the common ratio \( k \).
Given \( k = \frac{a_1}{a_2} \), we can write:

• \( a_2 = \frac{a_1}{k} \)
• \( a_3 = \frac{a_2}{k} = \frac{a_1}{k^2} \)
• \( a_4 = \frac{a_3}{k} = \frac{a_1}{k^3} \)

and so forth.
This pattern allows you to generalize the nth term of the sequence as \( a_n = \frac{a_1}{k^{n-1}} \). This structured approach helps in handling more complex problems involving geometric sequences efficiently.

Mathematical Proof

Mathematical proofs involve logical steps to show that a specific statement is universally true. In our exercise, the goal is to prove \( \frac{a_1^n}{a_2^n} = \frac{a_1}{a_{n+1}} \) using known properties of the sequence.
First, identify and use the common ratio to express each sequence term:

• \( a_2 = \frac{a_1}{k} \)
• \( a_n = \frac{a_1}{k^{n-1}} \)
• \( a_{n+1} = \frac{a_n}{k} = \frac{a_1}{k^n} \)

Next, rewrite the expression \( \frac{a_1^n}{a_2^n} \) using \( a_2 = \frac{a_1}{k} \):

• \( \frac{a_1^n}{a_2^n} = \frac{a_1^n}{\big(\frac{a_1}{k}\big)^n} = \frac{a_1^n}{\frac{a_1^n}{k^n}} = k^n \)

Finally, verify that it equals \( \frac{a_1}{a_{n+1}} = k^n \). Thus, the statement is proven correct.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions using algebraic rules. In the provided solution, it involves using algebraic operations to rewrite and simplify terms involving the common ratio \( k \).
Start with the expression \( \frac{a_1^n}{a_2^n} \): given that \( a_2 = \frac{a_1}{k} \), we have:

• \( \frac{a_1^n}{a_2^n} = \frac{a_1^n}{\big(\frac{a_1}{k}\big)^n} \)
• Simplify to get \( \frac{a_1^n}{\frac{a_1^n}{k^n}} = k^n \)

Next, express \( k^n \) in terms of initial sequence terms:

• \( k^n =\frac{a_1}{a_{n+1}} \)

This shows the power of breaking down complex expressions into manageable parts, ensuring clarity at each step. Algebraic manipulation is vital for solving many types of mathematical problems, making it an essential skill for students.