Answer
\[
\boxed{ y_k = \frac{k(k+1)(2k+1)}{6} }.
\]
Work Step by Step
First, note the recursive definition:
\[
\begin{cases}
y_1 = 1, \\
y_k = y_{k-1} + k^2 \quad \text{for } k \ge 2.
\end{cases}
\]
**Step 1: Unfold (iterate) the recursion**
For \(k \ge 2\),
\[
y_k
= y_{k-1} + k^2
= \bigl(y_{k-2} + (k-1)^2\bigr) + k^2
= \dots
= y_1 + 2^2 + 3^2 + \cdots + k^2.
\]
Since \(y_1 = 1\), this becomes
\[
y_k = 1 + 2^2 + 3^2 + \cdots + k^2.
\]
Notice that \(1 = 1^2\), so in fact:
\[
y_k = 1^2 + 2^2 + 3^2 + \cdots + k^2 = \sum_{m=1}^k m^2.
\]
**Step 2: Use the known formula for the sum of squares**
The sum of the first \(k\) squares is given by
\[
\sum_{m=1}^k m^2 = \frac{k(k+1)(2k+1)}{6}.
\]
Therefore, the explicit formula for \(y_k\) is:
\[
\boxed{ y_k = \frac{k(k+1)(2k+1)}{6} }.
\]
