Formula for the sequence c defined by Sequence b

[moporho]

Homework Statement

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

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

Homework Equations

C_n= $$\Sigma$$(i=1)^n b_i )

The Attempt at a Solution

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

C_n= $$\Sigma$$(i=1)^n b_i )=(n*1)^n +(n*1)^n +(n*1)^n +...

 

[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?

 

[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?

 

[Dick]

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

 

[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

 

[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)?

 

[moporho]

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

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

 

[Dick]

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

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

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

 

[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!

 

[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)?

 

[moporho]

C(2k+1)=k?

 

[Dick]

C(2k+1)=k?

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?

 

[moporho]

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

 

[Dick]

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

So that doesn't work either, right?

 

[moporho]

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

 

[Dick]

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

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.

 

[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?

 

[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).