satellite73
  • satellite73
the sequence is 2, 10, 30, 68,130, ... third differences is 6 so it is cubic what is the formula for each term?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i got the answer from my eyeballs, but the question is: what is the method for deriving the sequence?
anonymous
  • anonymous
what's the pattern in this sequence :/
experimentX
  • experimentX
man ... are you trying to guess by looking? did you check my last answer ... my second last drawing!!

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Zarkon
  • Zarkon
\[n^3+n\]
anonymous
  • anonymous
@satellite73 please tell what you got from your eyeballs
anonymous
  • 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
anonymous
  • anonymous
npower3+n
Zarkon
  • Zarkon
I used regression
anonymous
  • anonymous
@Zarkon expand...
anonymous
  • anonymous
please
Zarkon
  • Zarkon
I found the cubic regression formula for the above 'data' set
Zarkon
  • Zarkon
it is \[\hat{y}=x^3+x\]
anonymous
  • anonymous
hmmm you mean a system of equations?
Zarkon
  • 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 \[a=0,b=1,c=0,d=1,e=0\]
Zarkon
  • Zarkon
let \[x=\{1,2,3,4,5\}\] and \[y=\{2, 10, 30, 68,130\}\]
anonymous
  • anonymous
ok thanks. i think in this case maybe it was easier to guess, but in general not
Zarkon
  • Zarkon
sure
anonymous
  • 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 \(x=\{1,2,3,4\}\) \(y=\{2,10,30,68\}\)
Zarkon
  • Zarkon
sure
anonymous
  • anonymous
thnx
Zarkon
  • Zarkon
NP
experimentX
  • 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

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