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Algebra 9 Online
OpenStudy (anonymous):

Generate the first 5 terms of this sequence: f(1) = 1 and f(2) = 2, f(n) = f(n - 1) + f(n - 2), for n > 2. 1, 2, 3, 4, 5 1, 2, 3, 5, 7 1, 2, 3, 5, 8 1, 3, 5, 7, 9

OpenStudy (mathmale):

Lucky you! You're given the first 2 terms, so need only find the 3rd, 4th and 5th terms. \[f(1)=1~and~f(2)=2.\]

OpenStudy (anonymous):

its A right?

OpenStudy (mathmale):

Use the recursion formula given: \[f _{n}=f _{n-1}+f _{n-2}\]to find \[f _{3}.\]

OpenStudy (mathmale):

You'll need to show me how you got your answer before I'll say "yes" or "no" to that question. i'm most interested in helping you develop the skills necessary to solve these problems on your own.

OpenStudy (anonymous):

its says greater than two so wouldnt that be a?

OpenStudy (mathmale):

I'll be glad to respond after you've shown me (in writing) how you got your answer.

OpenStudy (anonymous):

I have no idea, that's why I'm asking.

OpenStudy (mathmale):

\[f _{n}=f _{n-1}+f _{n-2}\]can be interpreted in words as follows: "To find the nth term of this sequence, take the previous (n-1)th term and the one before that [the (n-2)th term and add them together. Thus, if you want the 3rd term, take the 2nd and the 1st terms and add them together. Please find the 3rd term, which is denoted by\[f _{3}=f _{2}+f _{1}\]

OpenStudy (mathmale):

You might want to look up "sequence" so as to be certain that you know what you're looking for. Sequences have terms. Your first terms, in this problem, are 1 and 2. You need to calculate the third, fourth and fifth terms, as before.

OpenStudy (anonymous):

f(3) is three right? how would i get f(4) or f(5)?

OpenStudy (anonymous):

found it out on my own, thanks a lot "Moderator"

OpenStudy (anonymous):

The answer is "C" for anyone who needs this in the future

OpenStudy (mathmale):

Very glad you were able to work this out on your own. Next time, however, please keep the answer to yourself; one of OpenStudy's major goals is to help members develop the skills they need to solve their own problems.

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