MHB CuRio$ty's question at Yahoo Answers regarding a linear homogeneous recursion

[MarkFL]

Here is the question:

Math help please! recursive formula?

consider the sequence defined by the following recursive formula and starting with A^1=3 and A^2=2
A^1=4A^N-1 +A^N-2
A)list the next four terms of the sequence
b) find A^g
NOTE: ALL SUBSCRIPTS ARE SUPPOSE TO BE LOCATED BELOW THE "A"
please show me how this is done. would liketo learn. thanks in advance!

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello CuRio\$!ty,

We are (presumably) given the linear homogeneous recursion:

$$A_{n}=4A_{n-1}+A_{n-2}$$

where $$A_1=3,\,A_2=2$$

a) List the next four terms of the sequence.

For this we may simply use the recursive algorithm:

$$A_3=4A_2+A_1=4\cdot2+3=11$$

$$A_4=4A_3+A_2=4\cdot11+2=46$$

$$A_5=4A_4+A_3=4\cdot46+11=195$$

$$A_6=4A_5+A_4=4\cdot195+46=826$$

b) Find $A_n$.

To find the closed form, we find the roots of the associated characteristic equation:

$$r^2-4r-1=0$$

The quadratic formula gives us:

$$r=2\pm\sqrt{5}$$

Hence, the close form is:

$$A_n=c_1\left(2+\sqrt{5} \right)^n+c_2\left(2-\sqrt{5} \right)^n$$

Using the initial values, we may determine the parameters $c_i$:

$$A_1=c_1\left(2+\sqrt{5} \right)+c_2\left(2-\sqrt{5} \right)=3$$

$$A_2=c_1\left(2+\sqrt{5} \right)^2+c_2\left(2-\sqrt{5} \right)^2=2$$

These equations may be written:

$$2\left(c_1+c_2 \right)+\sqrt{5}\left(c_1-c_2 \right)=3$$

$$9\left(c_1+c_2 \right)+4\sqrt{5}\left(c_1-c_2 \right)=2$$

Multiplying the first equation by -4 and adding to the second, we obtain:

$$c_1+c_2=-10$$

Multiplying the first equation by 9 and the second by -2 and adding we obtain:

$$c_1-c_2=\frac{23}{\sqrt{5}}$$

Adding together these last two equations, we get:

$$2c_1=\frac{23}{\sqrt{5}}-10\implies c_1=\frac{23}{2\sqrt{5}}-5$$

and so:

$$c_2=-\left(\frac{23}{2\sqrt{5}}+5 \right)$$

Thus, the closed form for the sequence is:

$$A_n=\left(\frac{23}{2\sqrt{5}}-5 \right)\left(2+\sqrt{5} \right)^n-\left(\frac{23}{2\sqrt{5}}+5 \right)\left(2-\sqrt{5} \right)^n$$

$$A_n=\frac{\sqrt{5}}{10}\left(\left(23-10\sqrt{5} \right)\left(2+\sqrt{5} \right)^n-\left(23+10\sqrt{5} \right)\left(2-\sqrt{5} \right)^n \right)$$