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 2 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?
 2 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?

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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0i got the answer from my eyeballs, but the question is: what is the method for deriving the sequence?

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.0what's the pattern in this sequence :/

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0man ... are you trying to guess by looking? did you check my last answer ... my second last drawing!!

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.0@satellite73 please tell what you got from your eyeballs

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0sequence 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

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.3I found the cubic regression formula for the above 'data' set

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0hmmm you mean a system of equations?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.3A 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 \[a=0,b=1,c=0,d=1,e=0\]

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.3let \[x=\{1,2,3,4,5\}\] and \[y=\{2, 10, 30, 68,130\}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0ok thanks. i think in this case maybe it was easier to guess, but in general not

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0oh, 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 \(x=\{1,2,3,4\}\) \(y=\{2,10,30,68\}\)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0Here is another one \[ 2 + \sum_{i = 1}^{n1}\left( 8 + \sum_{j=1}^{i1} 12 + (j1)6\right)\] Mathematica code .. 2 + Sum[8 + Sum[12 + (i  1) 6, {i, 1, j  1}], {j, 1, n  1}] though WA doesn't like it
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