Answer
\[
\boxed{s_k = k^2 + k + 3.}
\]
Work Step by Step
We are given the recurrence
\[
s_k \;=\; s_{k-1} \;+\; 2k,
\quad\text{for } k \ge 1,
\quad\text{with } s_0 = 3.
\]
A convenient way to guess a closed‐form is to **unroll** (iterate) the definition:
\[
\begin{aligned}
s_k
&= s_0 + 2\cdot 1 + 2\cdot 2 + 2\cdot 3 + \cdots + 2\cdot k \\
&= 3 + 2\sum_{j=1}^k j.
\end{aligned}
\]
We know the standard formula for the sum of the first \(k\) positive integers:
\[
\sum_{j=1}^k j \;=\; \frac{k(k+1)}{2}.
\]
Hence,
\[
2 \sum_{j=1}^k j
= 2 \cdot \frac{k(k+1)}{2}
= k(k+1).
\]
Putting this back into the expression for \(s_k\),
\[
s_k
= 3 \;+\; k(k+1)
= k^2 + k + 3.
\]
