# Thread: Basic Recurrence Relation Problem

1. ## Basic Recurrence Relation Problem

A factory makes custom sports cars at an increasing rate. In the first month only one car
is made, in the second month two cars are made, and so on, with n cars made in the nth
month.
(a) Set up a recurrence relation for the number of cars produced in the first n months by

this factory.
(b) How many cars are produced in the first year?

a. I think this would be an = an-1 + 1, but something tells me it might be an = an-1 + n, I am not sure.
b. Would this just be 1+2+3+4+5+6+7+8+9+10+11+12? Or, plugging in 12 for the above equation?

Thanks.

2. ## Re: Basic Recurrence Relation Problem

Hi Walshy,

Let's make a little diagram that we can check all of our answers with.

 Month Cars Made This month Cares Made to Date 1 1 1 2 2 3 3 3 6 4 4 10 5 5 15 ... ... ... n an Sn

A. Following the pattern, Sn = Sn-1+an. Yet, an = n, so we have that the number of cars made to date in the nth month is Sn = Sn-1+n. But Sn-1 = Sn-2 + n-1, so Sn = Sn-2 +(n-1) + n. Thus, continuing its recursive definition, Sn = Sn-3 + (n-2) + (n-1) + n. Continuing this until we get down to the first month, this is Sn = 1 + 2 + ... + n, which can be shown to equal (by the sum of an arithmetic sequence):
Sn = $\frac{n(n+1)}{2}$
With a quick check of out table, we can see that this gives us the third entry in every row.
$\frac{1*2}{2} = 1$
$\frac{2*3}{2} = 3$
$\frac{3*4}{2} = 6$
...

B. After the first year (n = 12), we can either substitute n=12 into the formula above or do 1 + 2 + ... + 12. Either way, we get 78 cars produced.