Answer
1. $ P_k \;=\; P_{k-1} + k,\quad P_1 = 2$
2. $P_n = 1+\frac{n(n+1)}{2} = \frac{n^2 + n + 2}{2}$
Work Step by Step
We want the maximum number of regions \(P_n\) into which \(n\) lines can divide the plane. It is well‐known (and can be proved by induction) that:
\[
\boxed{P_n \;=\; 1 \;+\; \frac{n(n+1)}{2} \;=\; \frac{n^2 + n + 2}{2}.
}
\]
Below is how we arrive at this formula via:
1. A **recurrence relation** in part (a).
2. An **explicit formula** in part (b).
---
## (a) Recurrence Relation
- With **\(n-1\) lines**, assume we have the maximum number of regions \(P_{n-1}\).
- When we add the \(n\)th line, in order to maximize the number of new regions, that new line should intersect all the existing \(n-1\) lines in distinct points (no three lines meeting at the same point).
Hence the \(n\)th line is cut into \(n\) segments by those \(n-1\) intersection points.
Each of those \(n\) segments will divide an existing region into two, thereby creating \(n\) **new** regions.
Therefore, the recurrence is:
\[
\boxed{P_n \;=\; P_{n-1} \;+\; n,\quad \text{for } n \ge 2,
\quad\text{with } P_1 = 2.}
\]
(We know \(P_1 = 2\) because a single line divides the plane into 2 regions.)
---
## (b) Guess (and Confirm) an Explicit Formula
Observe that the recurrence
\[
P_n = P_{n-1} + n
\]
suggests that \(P_n\) is a partial sum of the integers from 1 up to \(n\), plus some constant. Indeed:
\[
P_n
= P_1 + \bigl(2 + 3 + \cdots + n\bigr).
\]
Since \(P_1=2\), we write
\[
P_n
= 2 + \sum_{k=2}^n k
= 2 + \Bigl(\sum_{k=1}^n k - 1\Bigr)
= 2 + \left(\frac{n(n+1)}{2} - 1\right)
= 1 + \frac{n(n+1)}{2}.
\]
Hence, the closed‐form formula is
\[
\boxed{P_n
= 1 + \frac{n(n+1)}{2}
= \frac{n^2 + n + 2}{2}.}
\]
This is the **maximum** number of regions determined by \(n\) lines in the plane, assuming no two lines are parallel and no three lines intersect at a single point.
