if anot=2007 and a[n+1]=1+a[n]/1-a[n],n greater than 0.then

To solve the recurrence relation given by \( a_0 = 2007 \) and \( a_{n+1} = 1 + \frac{a_n}{1 - a_n} \) for \( n > 0 \), we can analyze how the sequence behaves as we iterate through the values. Let's break this down step by step.

Understanding the Recurrence Relation

The formula for \( a_{n+1} \) is defined in terms of \( a_n \). This means that each term in the sequence depends on the previous term. The expression \( 1 + \frac{a_n}{1 - a_n} \) can be rewritten to help us understand its behavior better.

Rearranging the Formula

We can rewrite the recurrence relation as follows:

• Let \( x = a_n \).
• Then, \( a_{n+1} = 1 + \frac{x}{1 - x} \).

To simplify this, we can find a common denominator:

• Multiply both the numerator and the denominator by \( 1 - x \):
• So, \( a_{n+1} = \frac{(1 - x) + x}{1 - x} = \frac{1}{1 - x} \).

This means that \( a_{n+1} = \frac{1}{1 - a_n} \). This transformation is crucial for understanding how the sequence evolves.

Exploring the Sequence

Now, let’s compute a few terms to see how the sequence behaves:

• Starting with \( a_0 = 2007 \):
• Calculate \( a_1 = 1 + \frac{2007}{1 - 2007} = 1 + \frac{2007}{-2006} = 1 - \frac{2007}{2006} = \frac{-1}{2006} \).
• Next, calculate \( a_2 = 1 + \frac{a_1}{1 - a_1} = 1 + \frac{-\frac{1}{2006}}{1 + \frac{1}{2006}} = 1 + \frac{-1}{2007} = \frac{2006}{2007} \).
• Then, calculate \( a_3 = 1 + \frac{2006/2007}{1 - 2006/2007} = 1 + \frac{2006/2007}{1/2007} = 1 + 2006 = 2007 \).

Identifying the Pattern

From our calculations, we see that:

• \( a_0 = 2007 \)
• \( a_1 = -\frac{1}{2006} \)
• \( a_2 = \frac{2006}{2007} \)
• \( a_3 = 2007 \)

This indicates that the sequence is periodic with a period of 3. Specifically, it cycles through the values \( 2007, -\frac{1}{2006}, \frac{2006}{2007} \). Thus, for any \( n \), the value of \( a_n \) can be determined by \( n \mod 3 \).

Final Thoughts

To summarize, the sequence defined by the recurrence relation is periodic with a cycle of three values. Depending on the value of \( n \mod 3 \), you can determine the corresponding term in the sequence:

• If \( n \mod 3 = 0 \), then \( a_n = 2007 \).
• If \( n \mod 3 = 1 \), then \( a_n = -\frac{1}{2006} \).
• If \( n \mod 3 = 2 \), then \( a_n = \frac{2006}{2007} \).

This periodic behavior is a fascinating aspect of recurrence relations, showcasing how complex sequences can emerge from simple rules.