Answer
See explanation
Work Step by Step
Below is a concise proof by mathematical induction that the sequence defined by
\[
x_k \;=\; 3\,x_{k-1} + k,
\quad k \ge 2,
\quad \text{with} \quad x_1 = 1,
\]
is given by the closed‐form formula
\[
\boxed{
x_k
= \frac{3^{k+1} - 2k - 3}{4}.
}
\]
---
## 1. Base Case \((k=1)\)
We need to check that the formula matches the initial condition \(x_1 = 1\). Plugging \(k=1\) into the closed form:
\[
x_1
= \frac{3^{1+1} - 2\cdot1 - 3}{4}
= \frac{3^2 - 2 - 3}{4}
= \frac{9 - 2 - 3}{4}
= \frac{4}{4} = 1.
\]
This matches the given \(x_1 = 1\). Hence, the base case is satisfied.
---
## 2. Inductive Step
**Inductive Hypothesis**: Suppose for some integer \(k \ge 2\), the formula holds for \(k-1\). In other words,
\[
x_{k-1}
= \frac{3^{(k-1)+1} - 2(k-1) - 3}{4}
= \frac{3^k - 2(k-1) - 3}{4}.
\]
**Goal**: Prove that
\[
x_k
= \frac{3^{k+1} - 2k - 3}{4}.
\]
**Proof**: By the recursion,
\[
x_k = 3\,x_{k-1} + k.
\]
Using the inductive hypothesis for \(x_{k-1}\),
\[
x_k
= 3 \Bigl(\tfrac{3^k - 2(k-1) - 3}{4}\Bigr) + k
= \frac{3\bigl(3^k - 2(k-1) - 3\bigr)}{4} + k.
\]
Distribute the 3 in the numerator:
\[
3\bigl(3^k - 2(k-1) - 3\bigr)
= 3^{k+1} - 6(k-1) - 9.
\]
Hence,
\[
x_k
= \frac{3^{k+1} - 6(k-1) - 9}{4} + k.
\]
Write \(k\) as \(\tfrac{4k}{4}\) to combine over a common denominator:
\[
x_k
= \frac{3^{k+1} - 6(k-1) - 9}{4}
+ \frac{4k}{4}
= \frac{3^{k+1} - 6(k-1) - 9 + 4k}{4}.
\]
Now simplify the expression inside the numerator:
\[
-6(k-1)
= -6k + 6,
\]
so
\[
3^{k+1} \;-\; 6(k-1) \;-\; 9 \;+\; 4k
= 3^{k+1} \;-\; 6k + 6 \;-\; 9 \;+\; 4k
= 3^{k+1} + (-6k + 4k) + (6 - 9)
= 3^{k+1} - 2k - 3.
\]
Therefore,
\[
x_k
= \frac{3^{k+1} - 2k - 3}{4},
\]
which matches exactly the closed‐form formula we needed to prove.
---
## 3. Conclusion
Since the base case holds and the inductive step has been verified, by the principle of mathematical induction, the formula
\[
\boxed{x_k = \frac{3^{k+1} - 2k - 3}{4}}
\]
is correct for all integers \(k \ge 1\) and satisfies both the recurrence \(x_k = 3\,x_{k-1} + k\) (for \(k\ge2\)) and the initial condition \(x_1 = 1\).
