Answer
\[
\boxed{
x_k \;=\; \frac{3^{\,k+1} - 3 - 2\,k}{4}
\quad=\quad \frac34\,3^k \;-\;\tfrac12\,k \;-\;\tfrac34.
}
\]
Work Step by Step
We have the recurrence
\[
x_k \;=\; 3\,x_{k-1} \;+\; k
\quad\text{for } k\ge2,
\quad\text{with}\quad x_1=1.
\]
A convenient way to solve this is to look for a closed form of the type \(A\,3^k + Bk + C\). However, you can also “unroll” the recursion or apply the standard method for linear recurrences. Below is a systematic derivation.
---
## 1. Solve the Homogeneous Part
First, consider the associated homogeneous recurrence
\[
x_k^{(h)} \;=\; 3\,x_{k-1}^{(h)}.
\]
Its general solution is
\[
x_k^{(h)} = C \, 3^k,
\]
for some constant \(C\).
---
## 2. Find a Particular Solution
Because the nonhomogeneous term is \(k\), we try a particular solution of the form
\[
x_k^{(p)} = A\,k + B.
\]
Substitute into the original recurrence \(x_k = 3\,x_{k-1} + k\):
\[
(Ak + B)
\;=\; 3\bigl[A(k-1) + B\bigr] \;+\; k
\;=\; 3\bigl(Ak - A + B\bigr) \;+\; k
\;=\; 3Ak - 3A + 3B + k.
\]
Compare coefficients of \(k\) and the constant terms:
- **Coefficient of \(k\):**
\(A\) on the left must equal \(3A + 1\) on the right. Hence
\[
A = 3A + 1
\;\Longrightarrow\;
-2A = 1
\;\Longrightarrow\;
A = -\tfrac12.
\]
- **Constant term:**
\(B\) on the left must equal \(-3A + 3B\) on the right. Substituting \(A = -\tfrac12\) gives
\[
B = -3\bigl(-\tfrac12\bigr) + 3B
\;\Longrightarrow\;
B = \tfrac32 + 3B
\;\Longrightarrow\;
B - 3B = \tfrac32
\;\Longrightarrow\;
-2B = \tfrac32
\;\Longrightarrow\;
B = -\tfrac34.
\]
Hence a particular solution is
\[
x_k^{(p)} \;=\; -\tfrac12\,k \;-\;\tfrac34.
\]
---
## 3. General Solution and Use the Initial Condition
Combining the homogeneous and particular solutions:
\[
x_k = x_k^{(h)} + x_k^{(p)}
= C \,3^k \;-\;\tfrac12\,k \;-\;\tfrac34.
\]
We now use \(x_1 = 1\) to solve for \(C\). When \(k=1\),
\[
x_1
= 1
= C\,3^1 \;-\;\tfrac12\cdot1 \;-\;\tfrac34
= 3C \;-\;\tfrac12 \;-\;\tfrac34
= 3C \;-\;\tfrac54.
\]
Thus
\[
3C = 1 + \tfrac54
= \tfrac94
\quad\Longrightarrow\quad
C = \tfrac34.
\]
So
\[
x_k
= \tfrac34\,3^k \;-\;\tfrac12\,k \;-\;\tfrac34.
\]
We can also rewrite this more compactly. Notice
\[
\tfrac34\,3^k = \frac{3^{k+1}}{4},
\quad
-\tfrac12\,k = -\frac{2k}{4},
\quad
-\tfrac34 = -\frac{3}{4}.
\]
Hence
\[
x_k
= \frac{3^{k+1}}{4} \;-\;\frac{2k}{4}\;-\;\frac{3}{4}
= \frac{\,3^{k+1} - 2k - 3\,}{4}.
\]
Either form is perfectly valid:
\[
\boxed{
x_k \;=\; \frac{3^{\,k+1} - 3 - 2\,k}{4}
\quad=\quad \frac34\,3^k \;-\;\tfrac12\,k \;-\;\tfrac34.
}
\]
