## satellite73 4 years ago the sequence is 2, 10, 30, 68,130, ... third differences is 6 so it is cubic what is the formula for each term?

1. anonymous

i got the answer from my eyeballs, but the question is: what is the method for deriving the sequence?

2. anonymous

what's the pattern in this sequence :/

3. experimentX

man ... are you trying to guess by looking? did you check my last answer ... my second last drawing!!

4. Zarkon

$n^3+n$

5. anonymous

6. anonymous

sequence is $$k^3+k$$ and yes, i guessed it, it wasn't hard you know it is cubic $2=1^3+1$$10=2^3+2$$30=3^3+3$ pattern is clear, i just wanted a snap method for getting it

7. anonymous

npower3+n

8. Zarkon

I used regression

9. anonymous

@Zarkon expand...

10. anonymous

11. Zarkon

I found the cubic regression formula for the above 'data' set

12. Zarkon

it is $\hat{y}=x^3+x$

13. anonymous

hmmm you mean a system of equations?

14. Zarkon

A polynomial of degree 4 can fit in the above data...do a 'linear' regression with $y=ax^4+bx^3+cx^2+dx+e$ you will get $a=0,b=1,c=0,d=1,e=0$

15. Zarkon

let $x=\{1,2,3,4,5\}$ and $y=\{2, 10, 30, 68,130\}$

16. anonymous

ok thanks. i think in this case maybe it was easier to guess, but in general not

17. Zarkon

sure

18. anonymous

oh, but i know it is a polynomial of degree 3, because the third differences are constant. i can ignore the last term then? and use $$x=\{1,2,3,4\}$$ $$y=\{2,10,30,68\}$$

19. Zarkon

sure

20. anonymous

thnx

21. Zarkon

NP

22. experimentX

Here is another one $2 + \sum_{i = 1}^{n-1}\left( 8 + \sum_{j=1}^{i-1} 12 + (j-1)6\right)$ Mathematica code .. 2 + Sum[8 + Sum[12 + (i - 1) 6, {i, 1, j - 1}], {j, 1, n - 1}] though W|A doesn't like it