How do I write 5+10+17+26 in sigma notation?

[123mathguy]

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

 

[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

 

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

 

[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.

 

[cks]

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

 

[cks]

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

where

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

 

[cks]

this is what I can do at most!

 

[matt grime]

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

 

[CRGreathouse]

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

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.)

 

[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.

 

[CRGreathouse]

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.

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.

 

[HallsofIvy]

In general, you can use "Newton's Difference Formula":
If you have a sequence {a_n} so that a₀= a, the first difference, a₁- a₀, is b, the "second difference" (subtract the first two first differences) is c, etc. then a_n= 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 x^k.