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Mathematics 19 Online
OpenStudy (anonymous):

The table below shows the values of f(n) for different values of n: n 1 2 3 4 5 6 f(n) 1 2 4 16 128 4096 Which recursive function best represents the values shown in the table? f(1) = 1, f(2) = 2, f(n) = f(n -3) + f(n - 2); n > 2 f(1) = 1, f(2) = 2, f(n) = f(n -3) f(n - 2); n > 2 f(1) = 1, f(2) = 2, f(n) = 2f(n -1) + f(n - 2); n > 2 f(1) = 1, f(2) = 2, f(n) = 2f(n -1) f(n - 2); n > 2

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

what do we know about a recursive function ... what does it tell us?

OpenStudy (amistre64):

notice that all the options differ only by the function rule that they state. so the rest of it is eye clutter. which rule defines the sequence?

OpenStudy (anonymous):

I'm not sure. :/

OpenStudy (amistre64):

tell me whats confusing you about the function rules ...

OpenStudy (anonymous):

I just dont know what to do to see if the values match up with the functions.

OpenStudy (amistre64):

f(n) = f(n -3) + f(n - 2) notice that there are 3 terms to deal with, and the rule for this one is just saying that the new terms is the sum of the 2 before it f(n) = f(n -3) f(n - 2), and this one is saying its the product of the 2 terms before it the rest are read in similar mathical nomenclatures

OpenStudy (amistre64):

one minor thing to note is that there is no f(0), so having some f(n-3) for n=3 is simply undefined and cannot be worked out. so those first 2 options are errors just by default.

OpenStudy (amistre64):

f(n) = 2f(n -1) + f(n - 2), let n=3 f(3) = 2f(3-1) + f(3 - 2) f(3) = 2f(2) + f(1) 4 = 2(2) + 1 ... is not right either so work the rule for the last option to verify

OpenStudy (anonymous):

Do I plug in 3 for n like you did?

OpenStudy (amistre64):

id hope so lol

OpenStudy (amistre64):

for n>2 means that n=3,4,5,6,...

OpenStudy (amistre64):

so testing it out for n=3 at lest lets us know its good for n=3

OpenStudy (anonymous):

Thanks! :) The last one is definitely the correct answer.

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