Answer
- Recurrence \(a_k = 2\,a_{k-1} + 3\,a_{k-2}\).
- Characteristic equation: \(t^2 - 2t - 3=0\).
- Roots: \(t=3\) or \(t=-1\).
- General form: \(a_k = A\,3^k + B\,(-1)^k\).
- With \(a_0=1, a_1=2\), we get \(a_k = \tfrac{3}{4}\,3^k + \tfrac{1}{4}\,(-1)^k.\)
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 \(a_k = 2\,a_{k-1} + 3\,a_{k-2}\)
### (a) Find all possible \(t\) so that \(t^n\) is a solution
1. Write the characteristic equation by setting \(a_k = t^k\):
\[
t^k = 2\,t^{k-1} + 3\,t^{k-2}.
\]
Divide by \(t^{k-2}\) (assuming \(t \neq 0\)):
\[
t^2 = 2\,t + 3.
\]
Hence the characteristic equation is
\[
t^2 - 2\,t - 3 = 0.
\]
2. Factor or use the quadratic formula:
\[
(t - 3)(t + 1) = 0
\quad\Longrightarrow\quad
t = 3 \quad\text{or}\quad t = -1.
\]
So the two possible values of \(t\) are \(3\) and \(-1\).
### (b) Given \(a_0 = 1\) and \(a_1 = 2\), find a closed‐form formula
From the theory, the general solution is
\[
a_k = A \, 3^k \;+\; B \, (-1)^k,
\]
for some constants \(A,B\). Use the initial conditions:
1. **\(k=0\)**:
\[
a_0 = 1 \;=\; A\,3^0 + B\,(-1)^0 = A + B.
\]
So \(A + B = 1.\)
2. **\(k=1\)**:
\[
a_1 = 2 \;=\; A\,3^1 + B\,(-1)^1 = 3A - B.
\]
So \(3A - B = 2.\)
Solve simultaneously:
- From \(A + B = 1\), we get \(B = 1 - A\).
- Substitute into \(3A - B = 2\):
\[
3A - (1 - A) = 2
\quad\Longrightarrow\quad
3A - 1 + A = 2
\quad\Longrightarrow\quad
4A = 3
\quad\Longrightarrow\quad
A = \tfrac{3}{4}.
\]
Then \(B = 1 - \tfrac{3}{4} = \tfrac{1}{4}.\)
Hence
\[
\boxed{a_k = \frac{3}{4}\,3^k \;+\; \frac{1}{4}\,(-1)^k,\quad k\ge 0.}
\]
