Answer
\[
\boxed{t_k = \left\lfloor \frac{k+1}{2} \right\rfloor.}
\]
Equivalently, for even \(k=2n\), \(t_k = n\); for odd \(k=2n+1\), \(t_k = n+1\).
Work Step by Step
**Solution Explanation**
We have the recurrence
\[
\begin{cases}
t_0 = 0,\\[4pt]
t_k = k \;-\; t_{k-1}, \quad \text{for all integers } k \ge 1.
\end{cases}
\]
---
## (a) Use Iteration to Guess an Explicit Formula
Let us compute the first several terms:
\(t_0 = 0\) (given).
\(t_1 = 1 - t_0 = 1 - 0 = 1.\)
\(t_2 = 2 - t_1 = 2 - 1 = 1.\)
\(t_3 = 3 - t_2 = 3 - 1 = 2.\)
\(t_4 = 4 - t_3 = 4 - 2 = 2.\)
\(t_5 = 5 - t_4 = 5 - 2 = 3.\)
\(t_6 = 6 - t_5 = 6 - 3 = 3.\)
\(t_7 = 7 - t_6 = 7 - 3 = 4.\)
\(t_8 = 8 - t_7 = 8 - 4 = 4.\)
\(t_9 = 9 - t_8 = 9 - 4 = 5.\)
We see a pattern where \(t_k\) increases by 1 every two steps:
\[
\begin{aligned}
k &= 0,1,2,3,4,5,6,7,8,9,\dots \\
t_k &= 0,1,1,2,2,3,3,4,4,5,\dots
\end{aligned}
\]
More precisely:
- For even \(k = 2n\), it looks like \(t_{2n} = n\).
- For odd \(k = 2n+1\), it looks like \(t_{2n+1} = n + 1\).
One succinct way to combine these cases is to observe:
\[
t_k = \left\lfloor \frac{k+1}{2} \right\rfloor.
\]
Indeed, if \(k\) is even \(\bigl(k=2n\bigr)\), then
\(\lfloor (k+1)/2 \rfloor = \lfloor (2n+1)/2 \rfloor = \lfloor n + 0.5 \rfloor = n.\)
If \(k\) is odd \(\bigl(k=2n+1\bigr)\), then
\(\lfloor (k+1)/2 \rfloor = \lfloor (2n+2)/2 \rfloor = \lfloor n+1 \rfloor = n+1.\)
Hence our “guess” for an explicit formula is:
\[
\boxed{ t_k = \left\lfloor \frac{k+1}{2} \right\rfloor, \quad k \ge 0. }
\]
Alternatively, one can write it as a piecewise function:
\[
t_k
= \begin{cases}
\dfrac{k}{2}, & \text{if \(k\) is even},\\[4pt]
\dfrac{k+1}{2}, & \text{if \(k\) is odd}.
\end{cases}
\]
Either form is correct.
---
## (b) Verification by Strong (or Simple) Induction
We prove the closed‐form
\[
t_k = \left\lfloor \frac{k+1}{2} \right\rfloor
\]
satisfies the recurrence \(t_k = k - t_{k-1}\) with \(t_0 = 0\).
### Base Case
- For \(k=0\):
The formula gives
\(\lfloor (0+1)/2 \rfloor = \lfloor 1/2 \rfloor = 0\),
which matches \(t_0 = 0\).
### Inductive Step
Assume for some integer \(k \ge 1\) that \(t_{k-1} = \lfloor k/2 \rfloor\). We need to show \(t_k = \lfloor (k+1)/2 \rfloor\). From the recurrence:
\[
t_k
= k \;-\; t_{k-1}
= k \;-\; \left\lfloor \frac{k}{2} \right\rfloor.
\]
Now consider whether \(k\) is even or odd:
- If \(k\) is **even**, say \(k = 2n\). Then \(\lfloor k/2 \rfloor = \lfloor 2n/2 \rfloor = n\). Hence
\[
t_k
= 2n - n
= n
= \left\lfloor \frac{2n+1}{2} \right\rfloor
= \left\lfloor \frac{k+1}{2} \right\rfloor.
\]
- If \(k\) is **odd**, say \(k = 2n+1\). Then \(\lfloor k/2 \rfloor = \lfloor (2n+1)/2 \rfloor = n\). So
\[
t_k
= (2n+1) - n
= n+1
= \left\lfloor \frac{(2n+1)+1}{2} \right\rfloor
= \left\lfloor \frac{k+1}{2} \right\rfloor.
\]
In both cases, \(t_k\) indeed equals \(\lfloor (k+1)/2 \rfloor\). This completes the inductive step.
