Problem 94 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a_{n+1}=\frac{1}{1-a_{n}}\) for \(n \geq 1\) and \(a_{3}=a_{1}\), then \(\left(a_{2001}\right)^{2001}=\) (A) \(\mathbb{1}\) (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

(A) 1

Step 1: Understanding the Sequence

We are given the recursive sequence defined by \( a_{n+1} = \frac{1}{1-a_n} \) for \( n \geq 1 \). This means each term is derived from the previous one using this formula. Additionally, we have \( a_3 = a_1 \) which indicates some periodicity in the sequence.

Step 2: Finding Terms for Pattern

Let's calculate the first few terms to identify any patterns. We know \( a_{3} = a_{1} \), suggesting a cycle. Start with an arbitrary \( a_1 \) as our sequence doesn't depend on its initial value because we are only asked about powers which can eliminate its dependency: 1. \( a_2 = \frac{1}{1-a_1} \)2. \( a_3 = \frac{1}{1-a_2} \) and since \( a_3 = a_1 \), there is a cycle of length 3.

Step 3: Cycle Determination

Given \( a_3 = a_1 \), continue with:- \( a_4 = \frac{1}{1 - a_3} \), which simplifies using our cycle assumption, - Continuing this pattern gives \( a_4 = a_2 \), \( a_5 = a_1 \), repeating every 3 terms.

Step 4: Analyzing \(a_{2001}\)

Since the sequence is periodic with period 3, let's determine the position of \( a_{2001} \). The sequence repeats every 3 terms:- Calculate \( 2001 \mod 3 \), which results in 2 because dividing 2001 by 3 gives a remainder of 2. Thus, \( a_{2001} = a_2 \).

Step 5: Calculating \((a_{2001})^{2001}\)

Since \( a_{2001} = a_2 \), and our calculation shows cycles repeat based solely on the iterative definition, given the problem's constraint, we ascertain - Compute \( (a_2)^{2001} \). Because of the repetitive and periodic nature, locating values are less consequential in predicting powers changing symmetric sequence properties.

Key Concepts

Periodicity

In sequences and series, periodicity refers to the repeating nature of the sequence after a set interval or number of terms. Identifying periodicity is crucial in solving problems with recursive sequences, especially when the problem does not provide initial values or when you're required to perform operations like multiple term manipulations. For the sequence defined by the exercise, we note that it repeats every three terms:

• We know that if the term after three places returns to its starting value, it suggests periodicity.
• The solution provides that since \( a_3 = a_1 \), it indicates that there is a cycle, repeating after every three steps.

This property makes solving sequences easier, since the behavior of any subsequent terms can be predicted by recognizing the cycle.

Cycle Determination

Cycle determination involves identifying the length of the repeating pattern in a sequence. This is crucial when working with recursive sequences because identifying the cycle informs us how the sequence behaves over an infinite number of terms. In the given exercise, the sequence has a cycle of length three:

• From the recursive formula \( a_{n+1} = \frac{1}{1-a_n} \), we calculate initial terms to uncover the cycle.
• Once we discover that \( a_3 = a_1 \), it is clear the sequence enters a cycle repeating every three terms.

Knowing that the sequence has a cycle of length three allows us to determine any term of the form \( a_{n} \) in the sequence quickly by using modulo operation, simplifying complex calculations.

Sequence Pattern

Recognizing sequence patterns involves observing and understanding the changes from one term to the next. This knowledge is fundamental when dealing with recursive sequences where each term depends on the preceding one. In the context of this sequence:

• The pattern arises from repeatedly applying the recursive rule to a starting term.
• By closely calculating subsequent terms, it becomes apparent that the sequence stabilizes into a periodic pattern after a few iterations.

Understanding this pattern helps in predicting future terms without computing each one individually, saving time and simplifying solutions.

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers wrap-around upon reaching a certain value, known as the modulus. Its application is especially useful in finding term positions within a repeating sequence. In this exercise:

• We use modular arithmetic to determine the position of \( a_{2001} \) within the periodic sequence.
• By calculating \( 2001 \mod 3 \), we find that the remainder is 2, which tells us that the 2001st term corresponds to \( a_2 \) in the cycle.

This application demonstrates how modular arithmetic can turn seemingly large and intimidating problems into manageable calculations by leveraging periodicity.