MHB Jackin's question at Yahoo Answers regarding a linear recurrence

[MarkFL]

Here is the question:

Please help solving the recurrence relation question?

View attachment 2104

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello jackin,

1.) If the current year's number of sales is the average of the previous two year's sales, then we may state:

$$c_{n}=\frac{c_{n-1}+c_{n-2}}{2}=\frac{1}{2}c_{n-1}+\frac{1}{2}c_{n-2}$$

2.) To find the closed form, let's write the recurrence as:

$$2c_{n}-c_{n-1}-c_{n-2}=0$$

Thus, we find the characteristic equation is:

$$2r^2-r-1=(r-1)(2r+1)=0$$

Hence, the characteristic roots are:

$$r=-\frac{1}{2},\,1$$

And so the closed form is:

$$c_{n}=\alpha_1\left(-\frac{1}{2} \right)^n+\alpha_2(1)^n$$

We may simplify this as:

$$c_{n}=\alpha_1\left(-\frac{1}{2} \right)^n+\alpha_2$$

Now we may use the given initial values to determine the parameters.

$$c_{1}=-\frac{1}{2}\alpha_1+\alpha_2=70$$

$$c_{2}=\frac{1}{4}\alpha_1+\alpha_2=100$$

Multiplying the second equation by 2 and adding to the first, there results:

$$3\alpha_2=270\implies \alpha_2=90$$

And using this value for $\alpha_2$ in the first equation, we find:

$$\alpha_1=40$$

And so the closed form is:

$$c_{n}=40\left(-\frac{1}{2} \right)^n+90$$