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

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) 1 (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The answer is C) 0.

Step 1: Understanding the Sequence

First, let's comprehend the sequence given by the recurrence relation. We know that: \[ a_{n+1} = \frac{1}{1-a_{n}} \] where \( n \geq 1 \). The sequence is determined once \( a_1 \) is known.

Step 2: Deduce Pattern in Sequence using Relation

We are also told \( a_3 = a_1 \). Observing this periodicity implies that the sequence is repeating itself every two terms, i.e., the sequence has a period of two.

Step 3: Determine Through Examples

Let us calculate the first few terms to confirm this periodicity. Assume \( a_1 = x \), then:\[ a_2 = \frac{1}{1-x} \]\[ a_3 = \frac{1}{1-a_2} = \frac{1}{1-\left(\frac{1}{1-x}\right)} = \frac{1-x}{x} = x \]Thus, \( a_1 = a_3 \) confirms the periodic nature.

Step 4: Generalize the Pattern

Since the sequence has a period of two, for any given \( n \),\[ a_n = x \text{ if } n \text{ is odd} \]\[ a_n = \frac{1}{1-x} \text{ if } n \text{ is even} \]

Step 5: Calculate \( a_{2001} \)

Since 2001 is an odd number, from the periodic pattern of the sequence, we have:\[ a_{2001} = a_1 \] and, from Step 3, let's assume \( a_1 = 0 \). Hence, \( a_{2001} = 0 \).

Step 6: Compute \( (a_{2001})^{2001} \)

If \( a_{2001} = 0 \), then:\[ (a_{2001})^{2001} = 0^{2001} = 0 \].

Key Concepts

Sequence Periodicity

Understanding sequence periodicity is crucial when studying sequences defined by recurrence relations. In the problem we are examining, the sequence is described by the recurrence relation \( a_{n+1} = \frac{1}{1-a_n} \). We are specifically told that \( a_3 = a_1 \), which suggests a repeating pattern. This repeating cycle is known as the sequence's period.

In simpler terms, a sequence is said to have periodicity if it repeats itself after a certain number of terms. Here, the periodicity is two, meaning the sequence cycles every two terms.

• When \( n \) is odd, \( a_n = a_1 \).
• When \( n \) is even, \( a_n = a_2 = \frac{1}{1 - a_1} \).

Recognizing this pattern greatly simplifies predicting future terms of the sequence. Instead of calculating terms indefinitely, we merely apply this periodic rule.

Recurrence Relation Pattern

Recurrence relations are equations that recursively define a sequence. They're essential in the study of sequences as they describe how each term in a sequence relates to its predecessor. In our exercise, the recurrence relation is given by \( a_{n+1} = \frac{1}{1-a_n} \). This formula tells us how to compute the next term based on the current term.

To find a recurrence relation pattern, we often calculate the first few terms and look for repetition or other notable regularities.
Here's how it works in our problem:

• Start with an initial value \( a_1 = x \).
• Compute \( a_2 = \frac{1}{1-x} \).
• Compute \( a_3 = \frac{1-x}{x} \) and observe that it's identical to \( x \) (hence the given \( a_3 = a_1 \)).

This relation forms a pattern due to the periodic nature, making it predictable and repetitive. Such recurrence patterns allow us to make deductions about any term without calculating each one individually.

Mathematical Induction

Mathematical induction is a powerful method used to prove statements, particularly those related to sequences and series. In problems involving recurrence relations, induction is used to verify whether a guess about the sequence holds true for all natural numbers.

Here's a simple breakdown of how it works in sequences:

• **Base Case**: Prove the statement is true for the initial term, such as \( n = 1 \).
• **Inductive Step**: Assume the statement holds for an arbitrary natural number \( k \), and then prove it true for \( k+1 \).

In the context of our specific problem, while the step-by-step solution doesn't detail explicit induction, it forms the backbone of proof in any formal inference process. When working with periodic sequences, induction helps prove the sequence's rules are defined correctly and hold consistently over the presumed period.