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Mathematics 8 Online
OpenStudy (anonymous):

Please help, will medal! Give a recursive definition for the set Y of all positive multiples of 5. That is, Y = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, ... }. Your definition should have a base case and a recursive part. 5 is in Y. If y is in Y, so is ___ .

OpenStudy (sasogeek):

duplicate http://openstudy.com/study#/updates/519ebd3ce4b0e9c70c329a34

OpenStudy (sasogeek):

Please read the OpenStudy terms and conditions http://openstudy.com/terms-and-conditions and OpenStudy Code of Conduct http://openstudy.com/code-of-conduct :)

OpenStudy (anonymous):

Yes i got it... this is not a test question

OpenStudy (loser66):

@Callisto , can you help her?please

OpenStudy (loser66):

@terenzreignz please, help

OpenStudy (anonymous):

i think the answer may have an exponent

OpenStudy (loser66):

show me your work, how to get exponent ?

OpenStudy (anonymous):

well the first part is 5 is in Y..

OpenStudy (loser66):

the whole work, please.

OpenStudy (anonymous):

because 5 is the multiple

OpenStudy (loser66):

multiple of what? your prof ask you start from n = 0 or n =1?, mine always start from n =0

OpenStudy (loser66):

ok, got what you mean, got what you want to go. (you have to speak out, so , i can know what you want, right? as I stated , there are many ways to come up, how can I know which way you want) continue, you are on the right way

OpenStudy (loser66):

from that , I guess, you start at n =1

OpenStudy (anonymous):

ok..

OpenStudy (loser66):

\[a_1 = 5\] \[a_2 = 10 = 5 + 1*5 = a_1 + (2-1)1*5 =a_1 +1*5\] \[a_3= 15 = 5 + 2*5 = a_1 + (3-1)*5 = a_1 + 2*5\] ................................. \[a_n= 5 + ( n-1)*5\] so the base case is 5 the recursive part is (n-1)*5 or 5(n-1) that's all I know,

OpenStudy (anonymous):

awesome thanks for your input :)

OpenStudy (loser66):

thanks God, at the end, I can get what you want.... hehehe... hey, please, next time, speak out what you want, if I can ,I will help, but need clue. ok? many many ways to solve, not just that.

OpenStudy (anonymous):

lol thanks :)

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