Answer
\[
\boxed{v_k = 3k - 2}
\]
Work Step by Step
We have the sequence \(\{v_k\}\) given by
\[
\begin{cases}
v_1 = 1,\\[4pt]
v_k \;=\; v_{\lfloor k/2 \rfloor} \;+\; v_{\lfloor (k+1)/2 \rfloor} \;+\; 2,
\quad \text{for all integers } k \ge 2.
\end{cases}
\]
---
## (a) Use Iteration (Compute First Few Terms) to Guess a Formula
Let’s calculate several terms:
\(v_1 = 1\) (given).
\(v_2 = v_{\lfloor 2/2 \rfloor} + v_{\lfloor 3/2 \rfloor} + 2
= v_1 + v_1 + 2
= 1 + 1 + 2
= 4.\)
\(v_3 = v_{\lfloor 3/2 \rfloor} + v_{\lfloor 4/2 \rfloor} + 2
= v_1 + v_2 + 2
= 1 + 4 + 2
= 7.\)
\(v_4 = v_{\lfloor 4/2 \rfloor} + v_{\lfloor 5/2 \rfloor} + 2
= v_2 + v_2 + 2
= 4 + 4 + 2
= 10.\)
\(v_5 = v_{\lfloor 5/2 \rfloor} + v_{\lfloor 6/2 \rfloor} + 2
= v_2 + v_3 + 2
= 4 + 7 + 2
= 13.\)
- \(v_6 = v_{\lfloor 6/2 \rfloor} + v_{\lfloor 7/2 \rfloor} + 2
= v_3 + v_3 + 2
= 7 + 7 + 2
= 16.\)
We see a pattern emerging:
\[
v_1 = 1,\;
v_2 = 4,\;
v_3 = 7,\;
v_4 = 10,\;
v_5 = 13,\;
v_6 = 16,\; \dots
\]
It appears each term increases by 3 from the previous term, starting from \(v_1 = 1\). This suggests an **arithmetic sequence** with common difference 3. Checking closely:
\[
v_k \;=\; 3k \;-\; 2.
\]
Indeed:
- For \(k=1\): \(3(1)-2=1.\)
- For \(k=2\): \(3(2)-2=6-2=4.\)
- For \(k=3\): \(3(3)-2=9-2=7.\)
- etc.
Hence we **guess** that
\[
\boxed{v_k = 3k - 2}.
\]
---
## (b) Verify the Formula by Strong (or Simple) Induction
We now prove that \(v_k = 3k - 2\) satisfies both the initial condition and the recursive relation.
### Base Case
For \(k=1\), our formula gives
\[
v_1 = 3(1) - 2 = 1,
\]
which matches the given \(v_1 = 1\). So the base case holds.
### Inductive Step
Assume that for **all** integers \(j\) with \(1 \le j < k\), we have \(v_j = 3j - 2\). We must show \(v_k = 3k - 2\).
From the recurrence:
\[
v_k
= v_{\lfloor k/2 \rfloor}
+ v_{\lfloor (k+1)/2 \rfloor}
+ 2.
\]
By the inductive hypothesis:
\[
v_{\lfloor k/2 \rfloor}
= 3\,\lfloor k/2 \rfloor - 2,
\quad
v_{\lfloor (k+1)/2 \rfloor}
= 3\,\lfloor (k+1)/2 \rfloor - 2.
\]
Hence,
\[
v_k
= \bigl[3\,\lfloor k/2 \rfloor - 2\bigr]
+ \bigl[3\,\lfloor (k+1)/2 \rfloor - 2\bigr]
+ 2.
\]
Combine like terms:
\[
v_k
= 3\,\lfloor k/2 \rfloor
+ 3\,\lfloor (k+1)/2 \rfloor
\;-\; 2 \;-\; 2
\;+\; 2
= 3\,\Bigl[\lfloor k/2 \rfloor + \lfloor (k+1)/2 \rfloor\Bigr]
\;-\; 2.
\]
It is a known fact (easy to check for even/odd \(k\)) that
\[
\lfloor k/2 \rfloor
\;+\; \lfloor (k+1)/2 \rfloor
\;=\; k.
\]
Thus,
\[
v_k
= 3\,k - 2.
\]
This completes the inductive step.
**Conclusion:** By the principle of mathematical induction, \(v_k = 3k - 2\) for all integers \(k \ge 1\).
