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Mathematics 18 Online
OpenStudy (anonymous):

3, 12, 21, 30, 39.... What is the arithmetic function of this sequence? I know we add 9 to it each time but I need to figure out the function. Kind of stuck...

OpenStudy (accessdenied):

What are your input values of your function, and what are your output values?

OpenStudy (anonymous):

I don't have any input values. The output values are as I indicated...3, 12, 21, 30, 39...etc.

OpenStudy (phi):

add 9 to it each time what do you do to 3 to get each number? add 9, then 2*9 , then 3*9...

OpenStudy (anonymous):

If I do f(n) = 3+9n this would work but only if things start at n=0

OpenStudy (accessdenied):

For the function, if we want it to represent this sequence, it probably should return the first value as 3 ( f(1) = 3 ), the next value as 12 ( f(2) = 12 ), then 21 ( f(3) = 21 ), and so on. So, these are our 'points.' We just have to figure out the general formula. Use the fact that you add 9 a different number of times to 3 to get to all of your function's output values...

OpenStudy (anonymous):

Doesn't the sequence start at n=1

OpenStudy (accessdenied):

Hm.. I think you could technically start it at either n=0 or n=1. It shouldn't matter too much, but I'm not sure if your teacher may have a preference that he/she used?

OpenStudy (anonymous):

She really didn't say...hmm..I guess I'll just try it and indicate that n starts at 0

OpenStudy (accessdenied):

K. If you wanted to move it to n=1 => f(1) = 3, you could just use a function shift: f(n) = 3 + 9n f(n-1) = 3 + 9(n-1) = 3 + 9n - 9 = -6 + 9n <-- call this the new f(n)

OpenStudy (accessdenied):

I think I would prefer the start at n=0, just because it is easier to see your first term. :)

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