Answer
- Recurrence \(b_k = 7\,b_{k-1} - 10\,b_{k-2}\).
- Characteristic equation: \(t^2 - 7t + 10=0\).
- Roots: \(t=5\) or \(t=2\).
- General form: \(b_k = A\,5^k + B\,2^k\).
- Specific \(b_0,b_1\) 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 \(b_k = 7\,b_{k-1} \;-\; 10\,b_{k-2}\)
### (a) Find all possible \(t\) so that \(t^n\) is a solution
Again, set \(b_k = t^k\) and derive the characteristic equation:
\[
t^k = 7\,t^{k-1} - 10\,t^{k-2}.
\]
Divide by \(t^{k-2}\):
\[
t^2 = 7\,t \;-\; 10,
\]
so
\[
t^2 - 7t + 10 = 0.
\]
Factor:
\[
(t - 5)\,(t - 2) = 0,
\]
so the roots are \(t=5\) or \(t=2\).
### (b) Suppose the sequence also satisfies \(b_0 = 2\) and \(b_1 = 1\) (for example). Find a closed‐form formula
(If the problem gives different values for \(b_1\), you would adjust accordingly. We'll assume \(b_1=1\) just as an illustration.)
The general solution is
\[
b_k = A\,5^k + B\,2^k.
\]
Use the initial conditions:
1. **\(k=0\)**:
\[
b_0 = 2 = A\,5^0 + B\,2^0 = A + B.
\]
So \(A + B = 2.\)
2. **\(k=1\)**:
\[
b_1 = 1 = A\,5^1 + B\,2^1 = 5A + 2B.
\]
So \(5A + 2B = 1.\)
Solve:
- From \(A + B = 2\), we have \(B = 2 - A\).
- Substitute into \(5A + 2B = 1\):
\[
5A + 2(2 - A) = 1
\quad\Longrightarrow\quad
5A + 4 - 2A = 1
\quad\Longrightarrow\quad
3A + 4 = 1
\quad\Longrightarrow\quad
3A = -3
\quad\Longrightarrow\quad
A = -1.
\]
Then \(B = 2 - (-1) = 3.\)
Hence
\[
\boxed{b_k = -\,5^k \;+\; 3\,2^k,\quad k\ge 0.}
\]
