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satellite73

  • 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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  1. satellite73
    • 2 years ago
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    i got the answer from my eyeballs, but the question is: what is the method for deriving the sequence?

  2. vikrantg4
    • 2 years ago
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    what's the pattern in this sequence :/

  3. experimentX
    • 2 years ago
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    man ... are you trying to guess by looking? did you check my last answer ... my second last drawing!!

  4. Zarkon
    • 2 years ago
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    \[n^3+n\]

  5. vikrantg4
    • 2 years ago
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    @satellite73 please tell what you got from your eyeballs

  6. satellite73
    • 2 years ago
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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

  7. bhaskarbabu
    • 2 years ago
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    npower3+n

  8. Zarkon
    • 2 years ago
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    I used regression

  9. satellite73
    • 2 years ago
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    @Zarkon expand...

  10. satellite73
    • 2 years ago
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    please

  11. Zarkon
    • 2 years ago
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    I found the cubic regression formula for the above 'data' set

  12. Zarkon
    • 2 years ago
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    it is \[\hat{y}=x^3+x\]

  13. satellite73
    • 2 years ago
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    hmmm you mean a system of equations?

  14. Zarkon
    • 2 years ago
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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 \[a=0,b=1,c=0,d=1,e=0\]

  15. Zarkon
    • 2 years ago
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    let \[x=\{1,2,3,4,5\}\] and \[y=\{2, 10, 30, 68,130\}\]

  16. satellite73
    • 2 years ago
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    ok thanks. i think in this case maybe it was easier to guess, but in general not

  17. Zarkon
    • 2 years ago
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    sure

  18. satellite73
    • 2 years ago
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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 \(x=\{1,2,3,4\}\) \(y=\{2,10,30,68\}\)

  19. Zarkon
    • 2 years ago
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    sure

  20. satellite73
    • 2 years ago
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    thnx

  21. Zarkon
    • 2 years ago
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    NP

  22. experimentX
    • 2 years ago
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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

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