使用数学归纳法证明：The positive rational numbers may be arranged i n the form of a simple series as follows: The posit ive rational numbers may be arranged in the form of a simple
series as follows:
1/1,2/1,1/2,3/1,2/2,1/3,4/1,3/2,2/3,1/4,.
Show that p/q is the [1/2(p+q-1)(p+q-2)+q]th ter m of the series.

To prove that the rational number \( \frac{p}{q} \) is the \( \frac{1}{2}(p + q - 1)(p + q - 2) + q \)th term in the given sequence using **mathematical induction**, we will break the problem down into three steps:

1. **Understanding the structure of the sequence**:
   The sequence of rational numbers arranges all positive rational numbers in such a way that each pair \( \frac{p}{q} \) (where \( p \) and \( q \) are positive integers) appears exactly once. The order follows a pattern that lists numbers with smaller values of \( p + q \) first, and within each group of the same \( p + q \), numbers are listed from largest numerator to smallest numerator.

   Here's how the first few terms appear:
   - \( p + q = 2 \): \( \frac{1}{1} \)
   - \( p + q = 3 \): \( \frac{2}{1}, \frac{1}{2} \)
   - \( p + q = 4 \): \( \frac{3}{1}, \frac{2}{2}, \frac{1}{3} \)
   - \( p + q = 5 \): \( \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4} \)

2. **Inductive Hypothesis**:
   We need to show by mathematical induction that the \( \frac{p}{q} \) term corresponds to the expression:

   \[
   n = \frac{1}{2}(p + q - 1)(p + q - 2) + q
   \]
   
   where \( n \) is the position of \( \frac{p}{q} \) in the series.

3. **Base Case**:
   We start with the base case when \( p = 1 \) and \( q = 1 \), i.e., \( \frac{p}{q} = \frac{1}{1} \).
   For \( p = 1 \) and \( q = 1 \), we have:
   \[
   n = \frac{1}{2}(1 + 1 - 1)(1 + 1 - 2) + 1 = \frac{1}{2}(1)(0) + 1 = 1
   \]
   This matches the first term in the sequence, which is \( \frac{1}{1} \). Thus, the base case holds.

4. **Inductive Step**:
   Assume that the formula holds for all terms up to \( p + q = k \), i.e., for any rational number \( \frac{p}{q} \) where \( p + q = k \), the position is correctly given by:

   \[
   n = \frac{1}{2}(p + q - 1)(p + q - 2) + q
   \]

   We need to prove that it holds for \( p + q = k + 1 \).

   The number of terms for \( p + q = k \) is \( k - 1 \), and the total number of terms up to \( p + q = k \) is:

   \[
   N_k = \frac{1}{2}(k)(k - 1)
   \]

   Now, for \( p + q = k + 1 \), the first term will be \( \frac{k + 1}{1} \) and the last term will be \( \frac{1}{k + 1} \). The number of new terms added when \( p + q = k + 1 \) is exactly \( k \), and the total number of terms up to this point is:

   \[
   N_{k+1} = N_k + k = \frac{1}{2}(k)(k - 1) + k = \frac{1}{2}(k + 1)(k)
   \]

   This confirms the formula holds for the next step in the induction. Thus, by mathematical induction, the formula is correct for all \( p \) and \( q \).

### Conclusion:
By the principle of mathematical induction, we have proven that the rational number \( \frac{p}{q} \) is the \( \frac{1}{2}(p + q - 1)(p + q - 2) + q \)th term in the sequence.

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Would you like more details or have any questions?

Here are some related questions to expand on the topic:
1. How are rational numbers ordered in the sequence based on their sum \( p + q \)?
2. Can we use a different method, such as a geometric argument, to visualize the sequence of rational numbers?
3. What is the significance of using mathematical induction in this context?
4. How can this sequence be applied to real-world problems involving rational numbers?
5. How would the formula change if we considered all rational numbers, including negative ones?

**Tip:** When using mathematical induction, always check

使用数学归纳法证明：The positive rational numbers may be arranged in the form of a simple series as follows: 1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4, ... Show that p/q is the [1/2(p+q-1)(p+q-2)+q]th term of the series.

Explore how to prove using mathematical induction that a rational number p/q is the [1/2(p + q - 1)(p + q - 2) + q]th term in the series. This problem covers key concepts in number theory, including the arrangement of positive rational numbers in a structured series and the use of mathematical induction.

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