Answer
\[
\boxed{t_k = \frac{3k^2 + 5k}{2}}.
\]
Work Step by Step
We are given the recursion
\[
t_k \;=\; t_{k-1} \;+\; 3k \;+\; 1,
\quad \text{for } k \ge 1,
\quad \text{with} \quad t_0 = 0.
\]
To guess a closed‐form formula, **unroll** the recursion:
\[
t_k
= t_0 + \sum_{i=1}^k \bigl(3i + 1\bigr).
\]
Since \(t_0 = 0\), it follows
\[
t_k = \sum_{i=1}^k \bigl(3i + 1\bigr).
\]
We can split this sum into two parts:
\[
\sum_{i=1}^k \bigl(3i + 1\bigr)
= 3 \sum_{i=1}^k i \;+\; \sum_{i=1}^k 1.
\]
We know the standard formulas:
\[
\sum_{i=1}^k i = \frac{k(k+1)}{2}
\quad\text{and}\quad
\sum_{i=1}^k 1 = k.
\]
Hence,
\[
3 \sum_{i=1}^k i
= 3 \cdot \frac{k(k+1)}{2}
= \frac{3k(k+1)}{2},
\]
and
\[
\sum_{i=1}^k 1 = k.
\]
Combining these:
\[
\sum_{i=1}^k \bigl(3i + 1\bigr)
= \frac{3k(k+1)}{2} + k
= \frac{3k(k+1) + 2k}{2}
= \frac{3k^2 + 3k + 2k}{2}
= \frac{3k^2 + 5k}{2}.
\]
Therefore,
\[
t_k
= \frac{3k^2 + 5k}{2}.
\]
