 vazelinahS

2021-08-20

Solve the recursion:

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

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
$\mathrm{¬}\left\{9{\alpha }_{1}\right\}+6{\alpha }_{2}=3$

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