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To solve this question, we are working on proving a statement \( P(n) \) using mathematical induction. The statement \( P(n) \) is given as:

\[
P(n): \quad \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1}
\]

For the basis step of mathematical induction, we need to show that \( P(1) \) is true. This will involve substituting \( n = 1 \) into the statement and checking if it holds.

### Step-by-Step Solution

1. **Basis Step \( P(1) \):**
   Substitute \( n = 1 \) in the equation:
   \[
   \frac{1}{1 \cdot 2} = \frac{1}{2}
   \]
   According to \( P(1) \), we expect this to be equal to \( \frac{1}{1+1} = \frac{1}{2} \).
   So, \( P(1) \) is true because both sides of the equation are equal.

2. **Examining Answer Choices:**
   - **\( P(1) \) is true:** This is correct based on our calculation.
   - **\( P(0) \) is true:** There is no information or basis for a \( P(0) \) in this context, as our sequence starts with \( n = 1 \).
   - **\( P(1) \rightarrow P(2) \) is true:** This would be part of the inductive step, which we are not required to check in the basis step. For the basis step, we only need \( P(1) \).
   - **\( \frac{1}{2} = \frac{1}{2} \):** This is true, but it is a generic statement that doesn't address the specific basis of \( P(1) \).
   - **\( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1} \) is true for \( n = 1 \):** This is the correct basis step statement, and it is true as we verified.
   - **\( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1} \) is true for some integer \( n \):** This statement is vague and does not focus on the basis step specifically.

### Conclusion
For the basis step, we should select:
- **\( P(1) \) is true**
- **\( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1} \) is true for \( n = 1 \)**

These are the appropriate selections to complete the basis step.

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Do you have any questions or need further clarification?

### Related Questions:
1. How would you proceed with the inductive step for proving \( P(n) \) for all \( n \)?
2. Why is \( P(0) \) not relevant in this context?
3. Can you explain why the basis step is critical in mathematical induction?
4. How would the proof change if the series started from \( n = 2 \) instead?
5. What would happen if \( P(1) \) was not true?

### Tip:
When using induction, always clearly separate the **basis step** (initial case) from the **inductive step** (assuming \( P(k) \) is true to prove \( P(k+1) \)).

Let P(n) be the statement that \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1} \).
We will have completed the basis step of the proof if we show that (Check all that apply):

Learn how to complete the basis step in a proof by mathematical induction for the series summation statement P(n): \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1} \). This problem involves key concepts in algebra and series, suitable for high school students in Grades 11-12.

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