Answer
See explanation
Work Step by Step
We have two sequences \(\{s_k\}\) and \(\{t_k\}\) each satisfying the same second‐order linear homogeneous recurrence with constant coefficients:
\[
\begin{aligned}
s_k &= 5\,s_{k-1} \;-\; 4\,s_{k-2},\\
t_k &= 5\,t_{k-1} \;-\; 4\,t_{k-2},
\end{aligned}
\quad \text{for all integers } k \ge 2.
\]
We want to show that the sequence \(\{2s_k + 3t_k\}\) also satisfies this same recurrence:
\[
2s_k + 3t_k
\;\stackrel{?}{=}\;
5\bigl(2s_{k-1} + 3t_{k-1}\bigr)
\;-\;
4\bigl(2s_{k-2} + 3t_{k-2}\bigr).
\]
---
## Proof
Let
\[
u_k \;=\; 2\,s_k \;+\; 3\,t_k.
\]
Then
\[
u_{k-1} = 2\,s_{k-1} + 3\,t_{k-1},
\quad
u_{k-2} = 2\,s_{k-2} + 3\,t_{k-2}.
\]
We check whether \(u_k\) satisfies the same recurrence:
\[
5\,u_{k-1} \;-\; 4\,u_{k-2}
= 5\bigl(2\,s_{k-1} + 3\,t_{k-1}\bigr)
\;-\; 4\bigl(2\,s_{k-2} + 3\,t_{k-2}\bigr).
\]
Distribute:
\[
= 2\bigl(5\,s_{k-1}\bigr)
+ 3\bigl(5\,t_{k-1}\bigr)
\;-\; 2\bigl(4\,s_{k-2}\bigr)
\;-\; 3\bigl(4\,t_{k-2}\bigr).
\]
Group the terms with \(s\) and \(t\):
\[
= 2\Bigl(5\,s_{k-1} - 4\,s_{k-2}\Bigr)
\;+\; 3\Bigl(5\,t_{k-1} - 4\,t_{k-2}\Bigr).
\]
But from the given recurrences,
\(\;5\,s_{k-1} - 4\,s_{k-2} = s_k\)\; and
\(\;5\,t_{k-1} - 4\,t_{k-2} = t_k.\)
Therefore,
\[
5\,u_{k-1} \;-\; 4\,u_{k-2}
= 2\,s_k \;+\; 3\,t_k
= u_k.
\]
Hence \(u_k\) satisfies
\[
u_k \;=\; 5\,u_{k-1} \;-\; 4\,u_{k-2},
\]
which is exactly the same second‐order linear recurrence. This completes the proof.
