vazelinahS

2021-08-20

Solve the recursion:

${A}_{1}=1.\text{}{A}_{2}=-1$

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

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 $a}_{n}={\alpha}_{1}{\left({r}_{0}\right)}^{n}+{\alpha}_{2}\left(n{r}_{0}\right\}{)}^{n$

Step 2

Given:

Given recurrence relation

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

Solution:

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

Therefore characteristic equation is${r}^{2}-5r+6=0$

Determinant:${b}^{2}-4ac={(-5)}^{2}-4\left(1\right)\left(6\right)=25-24=1>0$

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 $(r-3)(r-2)=0$

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)$\Rightarrow 3{\alpha}_{1}+2{\alpha}_{2}=1$

Now for${A}_{2}=-1$

$A}_{2}={\alpha}_{1}{\left(3\right)}^{2}+{\alpha}_{2}{\left(2\right)}^{2$

2)$\Rightarrow 9{\alpha}_{1}+4{\alpha}_{2}=-1$

By solving equation 1 and 2 we get

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

Concept Used:

To determine which general solution to use for a recurrence relation of degree 2, determinants can be helpful

from the quadratic equation

Case 1 if

we get two distinct roots

Case 2 if

we get one root with multiplicity 2

Step 2

Given:

Given recurrence relation

Solution:

Given

Therefore characteristic equation is

Determinant:

Since our determinant is greater than 0 we know that we get two distinct real roots

We can factor

So our roots are

Hence general solution is

Now we find

Now for

1)

Now for

2)

By solving equation 1 and 2 we get