# Finding the nth term,,

1. Apr 4, 2009

### remaan

1. The problem statement, all variables and given/known data

This type of questions seems to be easy, but a little confusing at the same time
If we are given the Sn of a series, and were asked to find an how to do that ?

2. Relevant equations

the limit of series and Sn is the same

3. The attempt at a solution

Maybe we should divide an and Sn ??

2. Apr 4, 2009

### Dick

What's the difference between S_n and S_(n+1)?

3. Apr 4, 2009

### remaan

I think that the only difference is one comes before the other, however they both have the same limits

4. Apr 4, 2009

### Dick

Maybe you'd better define what S_n means. I thought I knew what you were talking about. But your answer isn't reassuring.

5. Apr 4, 2009

### remaan

Sn is the partial sum of the nth term of a series .

6. Apr 4, 2009

### Dick

No, it's the partial sum of the first n terms in the series. a1+a2+a3+...+an. What would 'partial sum of an nth term' mean?

7. Apr 4, 2009

### remaan

Uha, and how to use to find an ?? When we know Sn as stated in the question ?

8. Apr 4, 2009

### Dick

I'll ask you again. Using slightly different words. What's S_(n+1)-S_(n)?

9. Apr 4, 2009

### remaan

I think that this is an right ?

SO do I have to find S(n+1) and subt. from S (n) ??

Is that what you mean ???

10. Apr 4, 2009

### Dick

That was supposed to be a hint, not the answer. I would say S_(n+1)-S_(n)=a_(n+1). You'll have to change it a little to get a_(n). But, yes, that's what I mean.

11. Apr 4, 2009

### remaan

mmm, ok what do you think of this
If the Sn = 3-n(2)^(-n )

I did what you suggested and I got that an = 2^(-n) ( n+2n+2)

Do you think that can be right ???

12. Apr 4, 2009

### Dick

That's not what I get. How did you work that out?

13. Apr 4, 2009

### remaan

an = Sn - S (n-1)

I found that Sn-1 = 3-(n-1)2^(-n+1)

Is that right, before I precede ???

14. Apr 4, 2009

### Dick

Looks ok to me.

15. Apr 4, 2009

### remaan

Ya, and that 's how I got the rest !!

I think I am fine "with this one "

Thanks alot Dick

16. Apr 4, 2009

### Dick

Ok, you're welcome! But I still don't get 2^(-n) ( n+2n+2).

17. Apr 4, 2009

### remaan

I got that by sub. Sn - S(n-1)

and then I took 2^(-n) as a common factor..

What do you think ?

18. Apr 4, 2009

### Dick

I think it's wrong. If you'd show the rest of your work maybe we could figure out why.

19. Apr 4, 2009

### remaan

Uha, sure :

here what I did :
Sn- S(n-1) = 3-n2(-n) - [ 3- n 2^(-n+1) - 2^ (-n+1) ]
3 we cancel and then we got what told you defore a while ,,

What do you think ?

20. Apr 4, 2009

### Dick

The '3's cancel, sure. I'm left with (n-1)*2^(-n+1)-n*2^(-n). If I factor 2^(-n) out I've got 2^(-n)*((n-1)*2-n). Right? Be more careful with the signs.

Last edited: Apr 4, 2009