Answer
- Recurrence \(c_k = c_{k-1} + 6\,c_{k-2}\).
- Characteristic equation: \(t^2 - t - 6=0\).
- Roots: \(t=3\) or \(t=-2\).
- General form: \(c_k = A\,3^k + B\,(-2)^k\).
- Specific \(c_0,c_1\) determine \(A,B\).
Work Step by Step
## General Background: Characteristic‐Equation Method
For a second‐order linear recurrence of the form
\[
x_k \;=\; A\,x_{k-1} \;+\; B\,x_{k-2},
\]
we look for solutions of the form \(x_k = t^k\). Substituting \(t^k\) into the recurrence and dividing by \(t^{k-2}\) (assuming \(t\neq 0\)) yields the **characteristic equation**:
\[
t^2 \;=\; A\,t \;+\; B,
\quad\text{i.e.}\quad
t^2 - A\,t - B = 0.
\]
Its two roots \(r_1,r_2\) (which might be equal) give the **general homogeneous solution**:
\[
x_k \;=\; \alpha\,r_1^k \;+\; \beta\,r_2^k.
\]
Then the initial conditions determine the constants \(\alpha,\beta\).
##The Recurrence \(c_k = c_{k-1} + 6\,c_{k-2}\)
### (a) Find all possible \(t\) so that \(t^n\) is a solution
Set \(c_k = t^k\). Then
\[
t^k = t^{k-1} + 6\,t^{k-2}.
\]
Divide by \(t^{k-2}\):
\[
t^2 = t + 6,
\]
so
\[
t^2 - t - 6 = 0.
\]
Factor:
\[
(t - 3)\,(t + 2) = 0,
\]
giving \(t=3\) or \(t=-2\).
### (b) Suppose the sequence also satisfies \(c_0 = 1\) and \(c_1 = 4\) (for instance). Find a closed‐form formula
The general solution is
\[
c_k = A\,3^k + B\,(-2)^k.
\]
Use the initial conditions:
1. **\(k=0\)**:
\[
c_0 = 1 = A\,3^0 + B\,(-2)^0 = A + B.
\]
So \(A + B = 1.\)
2. **\(k=1\)**:
\[
c_1 = 4 = A\,3^1 + B\,(-2)^1 = 3A - 2B.
\]
So \(3A - 2B = 4.\)
Solve:
- From \(A + B = 1\), we get \(B = 1 - A\).
- Substitute into \(3A - 2B = 4\):
\[
3A - 2(1 - A) = 4
\quad\Longrightarrow\quad
3A - 2 + 2A = 4
\quad\Longrightarrow\quad
5A - 2 = 4
\quad\Longrightarrow\quad
5A = 6
\quad\Longrightarrow\quad
A = \frac{6}{5}.
\]
Then \(B = 1 - \tfrac{6}{5} = -\tfrac{1}{5}.\)
Hence
\[
\boxed{c_k = \frac{6}{5}\,3^k \;-\; \frac{1}{5}\,(-2)^k,\quad k\ge 0.}
\]
