Sigma notaion help

1. Sep 30, 2007

123mathguy

can some one show me how to write
5+10+17+26 in sigma notation

2. Sep 30, 2007

rock.freak667

do you notice anything about those numbers? From 5 how do you get to 10? from 10 how to get to 17? from 17 how to get to 26? etc

3. Sep 30, 2007

CRGreathouse

One way would be
$$S=\{5,10,17,26\}$$
$$\sum_{x\in S}x$$

Another:
$$s_1=5,s_2=10,s_3=17,s_4=26$$
$$\sum_{k=1}^4s_k$$

Another:
$$\sum_{k=1}^4k^2+2k+2$$

4. Oct 1, 2007

cks

I can get the nth term that is

Tn=2+$$\sum_{k=1}^n2k+1$$

but I don't know how to get from the Tn to Sn.

Last edited: Oct 1, 2007
5. Oct 1, 2007

cks

How CRGreatHouse see that or how did he trasform from the nth term to the Sn term?

6. Oct 1, 2007

cks

$$\S_n=sum_{b=1}^nT_b$$

where

$$\T_b=2+sum_{k=1}^b2k+1$$

7. Oct 1, 2007

cks

this is what I can do at most!

8. Oct 1, 2007

matt grime

There is no 'unique' or 'canonical' way to do it. There are many ways. Yours (it appears to me) is fine.

9. Oct 1, 2007

CRGreathouse

I noticed that the second differences were constant, which means that a quadratic can be uniquely fitted to it. The sequence is 5, 10, 17, 26; the first differences are 5, 7, 9; the second differences are 2, 2. I then solved the system a+b+c=5, 4a+2b+c=10, 9a+3b+c=17. (I could have used 16a+4b+c=26 but didn't need it.)

Last edited: Oct 2, 2007
10. Oct 1, 2007

cks

I try some examples and see that when there's a difference of 2, then we can somewhat say that the nth term contains k^2,

well, but i fail to guess how you come out with the linear system, a+b+c=5, 4a+2b+c=10, 9a+3b+c=17.

Sorry, I have been trying hard to think about it.

11. Oct 2, 2007

CRGreathouse

I'm looking for a solution to y = ax^2 + bx + c with (x, y) = (1, 5), (2, 10), and (3, 17). Substitute and you get the linear system above.

12. Oct 2, 2007

HallsofIvy

Staff Emeritus
In general, you can use "Newton's Difference Formula":
If you have a sequence {an} so that a0= a, the first difference, a1- a0, is b, the "second difference" (subtract the first two first differences) is c, etc. then an= a+ bn+ (c/2)n(n-1)+ ... It looks a lot like a Taylor's series formula but you use n(n-1)(n-2)...(n-k) instead of xk.