Answer
See explanation
Work Step by Step
Below is a concise inductive proof that the sequence defined by
\[
s_k = s_{k-1} + 2k,
\quad k \ge 1,
\quad\text{with}\quad s_0 = 3,
\]
has the explicit formula
\[
\boxed{s_k = k^2 + k + 3}.
\]
---
## 1. Base Case \((k = 0)\)
We check that the formula holds for \(k=0\). By the proposed closed form,
\[
s_0 = 0^2 + 0 + 3 = 3.
\]
This matches the given initial condition \(s_0 = 3\). Hence, the base case is satisfied.
---
## 2. Inductive Step
**Inductive Hypothesis**: Assume for some integer \(k \ge 1\) that the formula holds at \(k-1\). In other words,
\[
s_{k-1} = (k-1)^2 + (k-1) + 3.
\]
**Goal**: Show that
\[
s_k = k^2 + k + 3.
\]
**Proof**: From the recursive definition,
\[
s_k
= s_{k-1} + 2k.
\]
Using the inductive hypothesis,
\[
s_k
= \Bigl[(k-1)^2 + (k-1) + 3\Bigr] + 2k.
\]
Now simplify:
\[
(k-1)^2 = k^2 - 2k + 1,
\]
so
\[
s_k
= \bigl(k^2 - 2k + 1\bigr) + (k-1) + 3 + 2k
= k^2 - 2k + 1 + k - 1 + 3 + 2k
= k^2 + \bigl(-2k + k + 2k\bigr) + (1 - 1 + 3)
= k^2 + k + 3.
\]
This completes the inductive step.
