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the sequence is 2, 10, 30, 68,130, ...
third differences is 6 so it is cubic
what is the formula for each term?
 one year ago
 one year ago
the sequence is 2, 10, 30, 68,130, ... third differences is 6 so it is cubic what is the formula for each term?
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.0
i got the answer from my eyeballs, but the question is: what is the method for deriving the sequence?
 one year ago

vikrantg4Best ResponseYou've already chosen the best response.0
what's the pattern in this sequence :/
 one year ago

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

vikrantg4Best ResponseYou've already chosen the best response.0
@satellite73 please tell what you got from your eyeballs
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.3
I found the cubic regression formula for the above 'data' set
 one year ago

ZarkonBest ResponseYou've already chosen the best response.3
it is \[\hat{y}=x^3+x\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
hmmm you mean a system of equations?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.3
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 \[a=0,b=1,c=0,d=1,e=0\]
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.0
ok thanks. i think in this case maybe it was easier to guess, but in general not
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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 \(x=\{1,2,3,4\}\) \(y=\{2,10,30,68\}\)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Here 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
 one year ago
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