Formula for the sequence c defined by Sequence b

1. Sep 26, 2008

moporho

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

Sequence b defined by bn=n(-1)^n, n≥1)

Find a formula for the sequence c defined by Cn= $$\Sigma$$(i=1)^n b_i )

2. Relevant equations
Cn= $$\Sigma$$(i=1)^n b_i )

3. The attempt at a solution

Should I solve for bi sequence first then solve for C?

Cn= $$\Sigma$$(i=1)^n b_i )=(n*1)^n +(n*1)^n +(n*1)^n +...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 26, 2008

Dick

Did you try writing down the first five or ten terms and see what it looks like? Can you describe it in words? You may find it easier to write separate formulas for the even and odd terms. Now can you prove it's correct? Probably by induction?

3. Sep 26, 2008

moporho

yes and this is what I came up with:

Cn=1(-1)^1 + 2(-1)^2 + 3(-1)^3 + 4(-1)^4 +5(-1)^5 + 6(-1)^6 + 7(-1)^7 + 8(-1)^8 + 9(-1)^9 + 10(-1)^10 ...n(-1)^n
= -1 + 2 + -3 + 4 + -5 + 6 + -7 + 8 + -9 + 10...n(-1)^n

so I think the answer should be {-1,-3,-5, -7 -9...} if n is negative
and {2,4,5,6,8,10...} if n is postive

Am I on the right path??? Does this look correct?

4. Sep 26, 2008

Dick

But you didn't add them up.
C1=-1
C2=-1+2
C3=-1+2-3
C4=-1+2-3+4
etc.

5. Sep 26, 2008

moporho

How is this for the final answer?
C1 = -1
C2 = 1
C3 = -2
C4 = 2
C5 = -3
C6 = 3
C7 = -4
C8 = 4
C9 = -5
C10 = 5

{-1,-2,-3, -4, -5,∞} if n is negative
{1, 2, 3, 4, 5,∞} if n is positive

6. Sep 26, 2008

Dick

The table is great. The description not so much. n is ALWAYS positive. It's either odd or even. Try writing it this way. If n is even it can be written as 2k (for some other integer k), if it's odd it can be written as 2k+1. So what are C(2k) and C(2k+1)?

7. Sep 26, 2008

moporho

C(2k) = even
C(2k+1) = odd

Thank you so much. It really helps to have someone walk you through. I appreciate it!

8. Sep 26, 2008

Dick

But you didn't say what C(2k) and C(2k+1) are!? You don't have a 'formula' yet.

9. Sep 26, 2008

moporho

C(2k) = 2, 4, 6, 8, 10,...n(-1)^n
C{2k + 1) = 1, 3, 5, 7, 9,...n(-1)^n
???

I am lost!

10. Sep 26, 2008

Dick

How about C(2k)=k? k=1 gives C(2)=1, k=2 gives C(4)=2. That fits with your table, yes? What is C(2k+1)?

11. Sep 26, 2008

moporho

C(2k+1)=k???

12. Sep 26, 2008

Dick

If that were correct then k=0 makes C(1)=0, k=1 makes C(3)=1, k=2 makes C(5)=2. If I refer to your table I see that the magnitude of those numbers is one too low, and the sign is wrong. Can you fix the formula?

13. Sep 26, 2008

moporho

C(2k + 1) = 2k + 1; 2k+1=3 gives C(2*3 + 1)=7, C(2*7 +1)=15

14. Sep 26, 2008

Dick

So that doesn't work either, right?

15. Sep 26, 2008

moporho

Now I am totally lost. It does not match the table.

16. Sep 26, 2008

Dick

It doesn't match the table because the formula is wrong. Can't you figure out how to fix the formula?? You want to match C(1)=(-1), C(3)=(-2), C(5)=(-3), etc. C(2k+1)=k and C(2k+1)=2k+1 are not the only choices available. Use your imagination! As I said, C(2k+1)=k is only a little wrong in two respects.

17. Sep 27, 2008

moporho

No I can not figure out the formula!! I am not seeing what you what and I do not understand! This is all I can come up with

C(2k+1)=-k
C(2(-1)+1 = -1
C(2(-2)+1 = -3
C(2(-3)+1 = -5
C(2(-4)+1 = -7

to get the

C(2(-3/2)+1) = -2
C(2(-5/2)+1) = -4
C(2(-7/2)+1) = -6

I do not understand! What does the formula look like?

18. Sep 27, 2008

Dick

I am saying that it is simplest to write TWO DIFFERENT formulas, one for even numbers and one for odd. If n is even and n=2k, write C(2k)=k (or C(n)=n/2). If n is odd and n=2k+1 write C(2k+1)=(-k-1) (or C(n)=(-n-1)/2).