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Math Help - Basic Recurrence Relation Problem

  1. #1
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    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.
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  2. #2
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    Nov 2012
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    Hanover, NH
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    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.
    Thanks from Walshy and wilhelm
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