Answer
See explanation
Work Step by Step
First, recall the piecewise definition given in Example 7.4.2 (slightly re‐labeled here for clarity):
\[
F(n)\;=\;
\begin{cases}
\dfrac{n}{2}, & \text{if \(n\) is even},\\[6pt]
-\dfrac{n-1}{2}, & \text{if \(n\) is odd}.
\end{cases}
\]
### (a) Checking the values for \(n=1,2,3,4\)
- **\(n=1\) (odd):**
\(F(1) = -\tfrac{1-1}{2} = -0 = 0.\)
- **\(n=2\) (even):**
\(F(2) = \tfrac{2}{2} = 1.\)
- **\(n=3\) (odd):**
\(F(3) = -\tfrac{3-1}{2} = -1.\)
- **\(n=4\) (even):**
\(F(4) = \tfrac{4}{2} = 2.\)
These match the intended pattern \(0,1,-1,2,\dots\), as described in the example.
---
### (b) A Single Formula Using the Floor Function
Notice that
\[
\left\lfloor \frac{n}{2} \right\rfloor
\;=\;
\begin{cases}
\tfrac{n}{2}, & \text{if \(n\) is even},\\
\tfrac{n-1}{2}, & \text{if \(n\) is odd}.
\end{cases}
\]
Moreover, \((-1)^n\) is \(+1\) for even \(n\) and \(-1\) for odd \(n\). Putting these observations together gives a compact, single‐expression formula:
\[
\boxed{F(n) \;=\; (-1)^n \,\bigl\lfloor \tfrac{n}{2} \bigr\rfloor.}
\]
- For **even** \(n\), \(\lfloor n/2 \rfloor = n/2\) and \((-1)^n = +1\), so \(F(n) = n/2\).
- For **odd** \(n\), \(\lfloor n/2 \rfloor = (n-1)/2\) and \((-1)^n = -1\), so \(F(n) = -\,(n-1)/2.\)
This exactly matches the original piecewise definition.
