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The problem asks to sum the probabilities for odd \( n \), \( \sum_{n=\text{odd}} p(n;t) \), to demonstrate that the crossover probability \( p(\text{crossover}) = \frac{1}{2} \left( 1 - e^{-2rt} \right) \), where \( r \) and \( t \) are parameters related to the system in question (such as rate and time).

To solve this, we will likely need to:

1. Consider that the function \( p(n;t) \) represents the probability for some event based on \( n \) and \( t \).
2. Use the summation over odd \( n \) values to represent the transition to the "crossover" state.
3. Apply appropriate summation techniques, probably leveraging properties of exponentials (e.g., a geometric series) to get the result in the form involving \( e^{-2rt} \).

Let’s break this down mathematically and explore the summation:

### Approach:
1. **Understand the function \( p(n;t) \):**
   - \( p(n;t) \) might have a known form, potentially involving a geometric or exponential series.
   - The term \( p(\text{crossover}) \) refers to summing probabilities over odd states, which suggests we are summing over specific discrete events.

2. **Sum over odd values of \( n \):**
   - We'll investigate the general form of \( p(n;t) \), apply this to odd \( n \), and show how this leads to the given formula.

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### Let’s start by clarifying: Would you like me to explain the steps mathematically based on a potential series form for \( p(n;t) \), or is there more context you can provide about the form of \( p(n;t) \)?

Sum the probabilities \(\sum_{n=odd} p(n;t)\) to show that \(p(crossover) = \frac{1}{2}(1 - e^{-2rt})\).

This solution shows how to sum the probabilities over odd terms in a series \(\sum_{n=odd} p(n;t)\) to derive the crossover probability formula \(p(crossover) = \frac{1}{2}(1 - e^{-2rt})\). Explore key concepts in probability, exponential decay, and series summation, suitable for advanced mathematics or physics students.

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