Solve the recursion: A1=1. A2=−1 Ak=5Ak−1−6Ak−2

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 ${\displaystyle a{x}^2+bx+c=0}$ the discriminant is ${\displaystyle b^2-4ac}$
Case 1 if ${\displaystyle b^2-4ac>0}$
we get two distinct roots ${\displaystyle r_1}$ and ${\displaystyle r_2}$ then general solution is ${\displaystyle a_n={\alpha }_1{\left(r_1\right)}^n+{\alpha }_2{\left(r_2\right)}^n}$
Case 2 if ${\displaystyle b^2-4ac=0}$
we get one root with multiplicity 2 ${\displaystyle r_0}$ then general solution is ${\displaystyle a_n={\alpha }_1{\left(r_0\right)}^n+{\alpha }_2\left(n{r}_0\right\}{)}^n}$
Step 2
Given:
Given recurrence relation
${\displaystyle A_k=5A_{k-1}-6A_{k-2},A_1=1,A_2=-1}$
Solution:
Given ${\displaystyle A_k=5A_{k-1}-6A_{k-2}}$
Therefore characteristic equation is ${\displaystyle r^2-5r+6=0}$
Determinant: ${\displaystyle b^2-4ac={(-5)}^2-4\left(1\right)\left(6\right)=25-24=1>0}$
Since our determinant is greater than 0 we know that we get two distinct real roots
We can factor ${\displaystyle r^2-5r+6=0}$ into ${\displaystyle (r-3)(r-2)=0}$
So our roots are ${\displaystyle r_1=3}$ and ${\displaystyle r_2=2}$
Hence general solution is ${\displaystyle A_k={\alpha }_1{\left(3\right)}^k+{\alpha }_2{\left(2\right)}^k}$
Now we find ${\displaystyle {\alpha }_1}$ and ${\displaystyle {\alpha }_2}$ by using given initial conditions
Now for ${\displaystyle A_1=1}$
${\displaystyle A_1={\alpha }_1{\left(3\right)}^1+{\alpha }_2{\left(2\right)}^1}$
1) ${\displaystyle \Rightarrow 3{\alpha }_1+2{\alpha }_2=1}$
Now for ${\displaystyle A_2=-1}$
${\displaystyle A_2={\alpha }_1{\left(3\right)}^2+{\alpha }_2{\left(2\right)}^2}$
2) ${\displaystyle \Rightarrow 9{\alpha }_1+4{\alpha }_2=-1}$
By solving equation 1 and 2 we get
${\displaystyle \mathrm{\neg }\left\{9{\alpha }_1\right\}+6{\alpha }_2=3}$