Answer
See explanation
Work Step by Step
Below is a concise proof by mathematical induction that the sequence defined by
\[
t_k = t_{k-1} + 3k + 1,\quad k \ge 1,\quad \text{with}\quad t_0 = 0,
\]
has the closed‐form solution
\[
\boxed{t_k = \frac{3k^2 + 5k}{2}}.
\]
---
## 1. Base Case \((k = 0)\)
We check that the formula holds for \(k=0\). Substituting \(k=0\) into the proposed closed form:
\[
t_0 = \frac{3\cdot 0^2 + 5\cdot 0}{2} = \frac{0}{2} = 0.
\]
This matches the given initial condition \(t_0 = 0\). Hence the base case is satisfied.
---
## 2. Inductive Step
**Inductive Hypothesis**: Assume that for some integer \(k \ge 1\) the formula holds for \(k-1\). In other words,
\[
t_{k-1} = \frac{3(k-1)^2 + 5(k-1)}{2}.
\]
**Goal**: Show that
\[
t_k = \frac{3k^2 + 5k}{2}.
\]
**Proof**: From the recursive definition,
\[
t_k
= t_{k-1} + 3k + 1.
\]
Using the inductive hypothesis for \(t_{k-1}\):
\[
t_k
= \frac{3(k-1)^2 + 5(k-1)}{2} + 3k + 1.
\]
Combine the terms over a common denominator of 2:
\[
t_k
= \frac{3(k-1)^2 + 5(k-1)}{2}
+ \frac{2\bigl(3k + 1\bigr)}{2}
= \frac{3(k-1)^2 + 5(k-1) + 6k + 2}{2}.
\]
Next, expand \((k-1)^2 = k^2 - 2k + 1\). Thus,
\[
3(k-1)^2
= 3\bigl(k^2 - 2k + 1\bigr)
= 3k^2 - 6k + 3,
\]
and
\[
5(k-1) = 5k - 5.
\]
Hence the numerator becomes
\[
3k^2 - 6k + 3 \;+\; 5k - 5 \;+\; 6k \;+\; 2
= 3k^2 + (-6k + 5k + 6k) + (3 - 5 + 2)
= 3k^2 + 5k + 0
= 3k^2 + 5k.
\]
Therefore,
\[
t_k
= \frac{3k^2 + 5k}{2},
\]
which is exactly the desired formula.
