vazelinahS

## Answered question

2021-08-20

Solve the recursion:

${A}_{k}=5{A}_{k-1}-6{A}_{k-2}$

### Answer & Explanation

Clara Reese

Skilled2021-08-21Added 120 answers

Step 1
Concept Used:
To determine which general solution to use for a recurrence relation of degree 2, determinants can be helpful
from the quadratic equation $a{x}^{2}+bx+c=0$ the discriminant is ${b}^{2}-4ac$
Case 1 if ${b}^{2}-4ac>0$
we get two distinct roots ${r}_{1}$ and ${r}_{2}$ then general solution is ${a}_{n}={\alpha }_{1}{\left({r}_{1}\right)}^{n}+{\alpha }_{2}{\left({r}_{2}\right)}^{n}$
Case 2 if ${b}^{2}-4ac=0$
we get one root with multiplicity 2 ${r}_{0}$ then general solution is ${a}_{n}={\alpha }_{1}{\left({r}_{0}\right)}^{n}+{\alpha }_{2}\left(n{r}_{0}\right\}{\right)}^{n}$
Step 2
Given:
Given recurrence relation

Solution:
Given ${A}_{k}=5{A}_{k-1}-6{A}_{k-2}$
Therefore characteristic equation is ${r}^{2}-5r+6=0$
Determinant: ${b}^{2}-4ac={\left(-5\right)}^{2}-4\left(1\right)\left(6\right)=25-24=1>0$
Since our determinant is greater than 0 we know that we get two distinct real roots
We can factor ${r}^{2}-5r+6=0$ into $\left(r-3\right)\left(r-2\right)=0$
So our roots are ${r}_{1}=3$ and ${r}_{2}=2$
Hence general solution is ${A}_{k}={\alpha }_{1}{\left(3\right)}^{k}+{\alpha }_{2}{\left(2\right)}^{k}$
Now we find ${\alpha }_{1}$ and ${\alpha }_{2}$ by using given initial conditions
Now for ${A}_{1}=1$
${A}_{1}={\alpha }_{1}{\left(3\right)}^{1}+{\alpha }_{2}{\left(2\right)}^{1}$
1) $⇒3{\alpha }_{1}+2{\alpha }_{2}=1$
Now for ${A}_{2}=-1$
${A}_{2}={\alpha }_{1}{\left(3\right)}^{2}+{\alpha }_{2}{\left(2\right)}^{2}$
2) $⇒9{\alpha }_{1}+4{\alpha }_{2}=-1$
By solving equation 1 and 2 we get
$\mathrm{¬}\left\{9{\alpha }_{1}\right\}+6{\alpha }_{2}=3$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?