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



Answered question


Solve the recursion:
A1=1. A2=1

Answer & Explanation

Clara Reese

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 ax2+bx+c=0 the discriminant is b24ac
Case 1 if b24ac>0
we get two distinct roots r1 and r2 then general solution is an=α1(r1)n+α2(r2)n
Case 2 if b24ac=0
we get one root with multiplicity 2 r0 then general solution is an=α1(r0)n+α2(nr0})n
Step 2
Given recurrence relation
Ak=5Ak16Ak2, A1=1, A2=1
Given Ak=5Ak16Ak2
Therefore characteristic equation is r25r+6=0
Determinant: b24ac=(5)24(1)(6)=2524=1>0
Since our determinant is greater than 0 we know that we get two distinct real roots
We can factor r25r+6=0 into (r3)(r2)=0
So our roots are r1=3 and r2=2
Hence general solution is Ak=α1(3)k+α2(2)k
Now we find α1 and α2 by using given initial conditions
Now for A1=1
1) 3α1+2α2=1
Now for A2=1
2) 9α1+4α2=1
By solving equation 1 and 2 we get

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