Question

Solve the recurrence relation an + an-1 – 6an-2 = 0 for n>=2 given that a0 = -1 and a1 = 8.​

Answer 1

Answer:

when n = 1 ,A1= 17a0+ 30 ,now A2= 17

Answer 2

We need to recall the following definition of the recurrence relation.

Recurrence relation: An equation that expresses the $n^{th}$ term of a sequence as a function of the preceding terms.

This problem is about the general solution of the recurrence relation.

Given:

$a_n+a_{n-1}-6a_{n-2}=0$  for $n\geq 2$

$a_0=-1$  and  $a_1=8$

We have,

$a_n+a_{n-1}-6a_{n-2}=0$

$a_n=-a_{n-1}+6a_{n-2}$

Here, the degree is $2$.

So, the characteristic equation is,

$x^{2} =-x+6$

$x^{2} +x-6=0$

$(x-2)(x+3)=0$

$x=2$  or  $x=-3$

Let's consider  $x_1=2$  and  $x_2=-3$

Then, the general solution of the homogeneous equation is,

$a_n=A_1x_1^n+A_2x_2^n$

$a_n=A_12^n+A_2(-3)^n$                        .......$(1)$

From the given initial conditions, we get

$a_0=-1\rightarrow\ n=0$

$a_0=A_12^0+A_2(-3)^0$

$-1=A_1+A_2$                                    .......$(2)$

$a_1=8\rightarrow\ n=1$

$8=A_12^1+A_2(-3)^1$

$8=2A_1-3A_2$                                   .......$(3)$

Solving equations $(2)$ and $(3)$ , we get

$A_1=1$  and  $A_2=-2$

Substitute the values in the equation $(1)$

Hence, the general solution of the recurrence relation is,

$a_n=+2^n-2(-3)^n$