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Mathematics 19 Online
OpenStudy (anonymous):

Generate the first 5 terms of this sequence: f(1) = 2 and f(2) = 3, f(n) = f(n - 1) + f(n - 2), for n > 2. 2, 3, 5, 7, 9 2, 3, 4, 7, 11 2, 3, 5, 9, 11 2, 3, 5, 8, 13

OpenStudy (anonymous):

Giving a medal for who ever can gelp me with this.

OpenStudy (anonymous):

@mathslover

mathslover (mathslover):

First of all, you are given with a function of n , such that : \(\bf{ f(n) = f(n-1) + f(n-2) }\) For f(1) , we will just put n =1 to find f(1) . Similarly, for f(2) , we will just put n =2 in f(n) to find f(2) . . .

mathslover (mathslover):

The first two terms are given to you, right?

mathslover (mathslover):

@A.Squad - Please respond so that I can help you. Else, it will not be possible for me to help you. Thanks.

OpenStudy (anonymous):

Yes, it is given to you.

mathslover (mathslover):

Right. So, can you find f(3) , f(4) and f(5) ?

OpenStudy (anonymous):

So, FN 3 is just FN 1 + FN 2?

mathslover (mathslover):

Yes, right!

OpenStudy (anonymous):

Then F4 is?

OpenStudy (anonymous):

F2 + F 3?

mathslover (mathslover):

Yes, good going.

OpenStudy (anonymous):

Okay,so what's up with F(n)?

mathslover (mathslover):

First of all, let us find f(3) f(3) = f(2) + f(1) = ?

OpenStudy (anonymous):

5

mathslover (mathslover):

Good. And similarly, can you find f(4) ?

OpenStudy (anonymous):

That's F2 + F3 so... 8?

mathslover (mathslover):

Right.

mathslover (mathslover):

Similarly find f(5) .

OpenStudy (anonymous):

F3 +F4 = 13?

mathslover (mathslover):

Correct!

OpenStudy (anonymous):

So F1 = 2, F2 = 3, F3 = 5, F4 = 8, F5 = 13?

mathslover (mathslover):

Very well done.

OpenStudy (anonymous):

Thanks so much, man...

OpenStudy (anonymous):

You're a life saver, lol

mathslover (mathslover):

I didn't do anything. You did the work - This is OpenStudy!

OpenStudy (anonymous):

lol you still taught me how to do it.

mathslover (mathslover):

:) Good Luck with your future studies.

OpenStudy (anonymous):

Thanks :)

OpenStudy (anonymous):

Let you know if there's anything.

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