satellite73
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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anonymous
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i got the answer from my eyeballs, but the question is: what is the method for deriving the sequence?
vikrantg4
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what's the pattern in this sequence :/
experimentX
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man ... are you trying to guess by looking? did you check my last answer ... my second last drawing!!
Zarkon
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\[n^3+n\]
vikrantg4
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@satellite73 please tell what you got from your eyeballs
anonymous
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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
bhaskarbabu
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npower3+n
Zarkon
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I used regression
anonymous
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@Zarkon expand...
anonymous
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please
Zarkon
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I found the cubic regression formula for the above 'data' set
Zarkon
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it is \[\hat{y}=x^3+x\]
anonymous
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hmmm you mean a system of equations?
Zarkon
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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
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let \[x=\{1,2,3,4,5\}\] and \[y=\{2, 10, 30, 68,130\}\]
anonymous
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ok thanks. i think in this case maybe it was easier to guess, but in general not
Zarkon
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sure
anonymous
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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
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sure
anonymous
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thnx
Zarkon
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NP
experimentX
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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