Answer
\[
\boxed{
w_k
= \begin{cases}
n^2, & \text{if } k=2n-1 \text{ (odd $k$)},\\[4pt]
n(n+1), & \text{if } k=2n \text{ (even $k$)}.
\end{cases}
}
\]
Equivalently,
\[
w_k
= \begin{cases}
\bigl(\tfrac{k+1}{2}\bigr)^2, & \text{if $k$ is odd},\\[4pt]
\tfrac{k}{2}\,\Bigl(\tfrac{k}{2}+1\Bigr), & \text{if $k$ is even}.
\end{cases}
\]
Work Step by Step
We have the sequence \(\{w_k\}\) defined by
\[
\begin{cases}
w_1 = 1,\quad w_2 = 2,\\[6pt]
w_k \;=\; w_{k-2} \;+\; k,\quad \text{for all integers } k \ge 3.
\end{cases}
\]
---
## (a) Use Iteration (Compute First Terms) to Guess a Formula
Let us compute several terms:
\(w_1 = 1\) (given).
\(w_2 = 2\) (given).
\(w_3 = w_1 + 3 = 1 + 3 = 4.\)
\(w_4 = w_2 + 4 = 2 + 4 = 6.\)
\(w_5 = w_3 + 5 = 4 + 5 = 9.\)
\(w_6 = w_4 + 6 = 6 + 6 = 12.\)
\(w_7 = w_5 + 7 = 9 + 7 = 16.\)
\(w_8 = w_6 + 8 = 12 + 8 = 20.\)
\(w_9 = w_7 + 9 = 16 + 9 = 25.\)
\(w_{10} = w_8 + 10 = 20 + 10 = 30.\)
Listing them out:
\[
\begin{aligned}
k &: 1,\;2,\;3,\;4,\;5,\;6,\;7,\;8,\;9,\;10,\;\dots\\
w_k &: 1,\;2,\;4,\;6,\;9,\;12,\;16,\;20,\;25,\;30,\;\dots
\end{aligned}
\]
Look for patterns:
- **Odd indices** \(k = 1,3,5,7,9,\dots\):
\(w_1 = 1,\; w_3=4,\; w_5=9,\; w_7=16,\; w_9=25,\dots\)
These are perfect squares: \(1=1^2,\;4=2^2,\;9=3^2,\;16=4^2,\;25=5^2,\dots\)
Indeed, if \(k = 2n - 1\) is odd, then \(w_k = n^2\).
- **Even indices** \(k = 2,4,6,8,10,\dots\):
\(w_2=2,\; w_4=6,\; w_6=12,\; w_8=20,\; w_{10}=30,\dots\)
Checking these values suggests \(w_{2n} = n(n+1)\).
Indeed,
\[
w_2 = 1\cdot 2 = 2,\quad
w_4 = 2\cdot 3 = 6,\quad
w_6 = 3\cdot 4 = 12,\quad
w_8 = 4\cdot 5 = 20,\quad
w_{10} = 5\cdot 6 = 30,\dots
\]
Hence a natural piecewise explicit formula is:
\[
\boxed{
w_k
= \begin{cases}
n^2, & \text{if }k = 2n - 1\text{ (i.e.\ $k$ odd)},\\[6pt]
n(n+1), & \text{if }k = 2n\text{ (i.e.\ $k$ even)}.
\end{cases}
}
\]
Equivalently, one can rewrite it in terms of \(k\) directly:
- If \(k\) is odd \(\bigl(k=2n-1\bigr)\), then \(n = \tfrac{k+1}{2}\) and
\(w_k = \bigl(\tfrac{k+1}{2}\bigr)^2.\)
- If \(k\) is even \(\bigl(k=2n\bigr)\), then \(n = \tfrac{k}{2}\) and
\(w_k = \tfrac{k}{2}\,\Bigl(\tfrac{k}{2} + 1\Bigr) = \tfrac{k(k+2)}{4}.\)
Either piecewise form is correct.
---
## (b) Verification by Strong (or Simple) Induction
We will prove the piecewise description
\[
w_{2n-1} = n^2,
\quad
w_{2n} = n(n+1),
\]
for all integers \(n \ge 1\). (This covers \(k\ge 1\).)
### Base Cases
- **\(k=1\)** (odd). Then \(n=1\) since \(1=2\cdot1-1\). We have \(w_1=1\). The formula says \(w_1 = 1^2 = 1\). Checks out.
- **\(k=2\)** (even). Then \(n=1\) since \(2=2\cdot1\). We have \(w_2=2\). The formula says \(w_2 = 1\cdot(1+1) = 2\). Checks out.
### Inductive Step
Suppose for all indices less than \(k\), the formula holds. We must show it holds for \(k\).
1. **Case 1: \(k\) is odd, say \(k=2n-1\).**
Then \(k-2=2n-3\) is also odd, so by the inductive hypothesis,
\[
w_{k-2}
= w_{2n-3}
= (n-1)^2.
\]
The recurrence says \(w_k = w_{k-2} + k\). Hence
\[
w_{2n-1}
= (n-1)^2 + (2n-1)
= n^2 - 2n + 1 + 2n - 1
= n^2,
\]
exactly as desired.
2. **Case 2: \(k\) is even, say \(k=2n\).**
Then \(k-2=2n-2\) is also even, so by the inductive hypothesis,
\[
w_{k-2}
= w_{2n-2}
= (n-1)\bigl((n-1)+1\bigr)
= (n-1)n.
\]
Again using \(w_k = w_{k-2} + k\),
\[
w_{2n}
= (n-1)n + (2n)
= n^2 - n + 2n
= n^2 + n
= n(n+1).
\]
That completes the even case.
By induction, the piecewise formula holds for all \(k \ge 1\). (And \(w_1=1\) is consistent with the initial condition.)
