Answer
\[
\boxed{p_k = 3^{k+1} - 7.}
\]
Work Step by Step
We are given the recursion
\[
p_k \;=\; p_{k-1} \;+\; 2 \cdot 3^k,
\quad \text{for } k \ge 2,
\quad \text{with} \quad p_1 = 2.
\]
To find a closed form, **unroll** (iterate) the recurrence:
\[
\begin{aligned}
p_k
&= p_{1} \;+\; 2 \cdot 3^2 \;+\; 2 \cdot 3^3 \;+\; \cdots \;+\; 2 \cdot 3^k \\
&= 2 \;+\; 2 \sum_{n=2}^k 3^n.
\end{aligned}
\]
Next, we evaluate the geometric sum:
\[
\sum_{n=2}^k 3^n
\;=\; \Bigl(\sum_{n=0}^k 3^n\Bigr)
\;-\; \bigl(3^0 + 3^1\bigr)
\;=\; \frac{3^{\,k+1} - 1}{3 - 1} - (1 + 3)
\;=\; \frac{3^{\,k+1} - 1}{2} - 4
\;=\; \frac{3^{\,k+1} - 9}{2}.
\]
Hence,
\[
2 \sum_{n=2}^k 3^n
= 2 \cdot \frac{3^{\,k+1} - 9}{2}
= 3^{\,k+1} - 9.
\]
Putting this back into the expression for \(p_k\),
\[
p_k
= 2 \;+\; \Bigl(3^{\,k+1} - 9\Bigr)
= 3^{\,k+1} \;-\; 7.
\]
