Answer
See explanation
Work Step by Step
We want to show, by mathematical induction, that the sequence defined by
\[
p_k \;=\; p_{k-1} + 2\cdot 3^k,
\quad k \ge 2,
\quad \text{with} \; p_1 = 2,
\]
is given by the explicit formula
\[
\boxed{p_k = 3^{k+1} \;-\; 7.}
\]
---
## 1. Base Case \((k = 1)\)
We check that the formula holds for \(k=1\). Plugging \(k=1\) into the proposed formula:
\[
p_1 = 3^{1+1} - 7 = 3^2 - 7 = 9 - 7 = 2.
\]
This matches the given initial condition \(p_1 = 2\). Hence the base case is satisfied.
---
## 2. Inductive Step
**Inductive Hypothesis**: Assume that for some integer \(k \ge 2\), the formula holds at \(k-1\); that is,
\[
p_{k-1} = 3^{(k-1)+1} - 7 = 3^k - 7.
\]
**To Prove**: Under this assumption, the formula also holds at \(k\), namely
\[
p_k = 3^{k+1} - 7.
\]
**Proof**: Using the recursive definition,
\[
p_k
\;=\; p_{k-1} + 2\cdot 3^k.
\]
By the inductive hypothesis \(p_{k-1} = 3^k - 7\). Substituting,
\[
p_k
= (3^k - 7) \;+\; 2\cdot 3^k
= 3^k + 2\cdot 3^k - 7
= 3^{k+1} - 7.
\]
This is exactly the desired form for \(p_k\).
