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Math Help - Number sequence

  1. #1
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    Number sequence

    5, 10, 17, 26, 37, ...

    I see the pattern, but I can't figure out the equation...any advice?
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  2. #2
    Member u2_wa's Avatar
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    Quote Originally Posted by jzellt View Post
    5, 10, 17, 26, 37, ...

    I see the pattern, but I can't figure out the equation...any advice?
    Hello jzellt:

    You might have noticed that difference between the numbers in sequence corresponds to the odd numbers sequence starting from 5.

    5,7,9,11,.....

    Now we can write it as 5+0,5+5,5+5+7,5+5+7+9,...
    We can use this formula for the addition of the odd numbers:
    \frac{n}{2}(2a+(n-1)*d)

    Lets plug in the values:
    \frac{n}{2}(2(5)+(n-1)*2)
    \frac{n-1}{2}(10+(n-2)*2) (because at n=1, there should be addition of zero i.e 5+0,5+5,....)
    Hence we have the equation: 5+\frac{n-1}{2}(8+2(n-1))

    Hope this helps
    Last edited by u2_wa; March 13th 2009 at 07:25 AM.
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  3. #3
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    Hello, jzellt!

    I "eyeballed" the solution . . .


    5, 10, 17, 26, 37, \hdots

    Take the differences of consecutive terms.
    Then take the differences of the differences ... and so on.

    . . \begin{array}{cccccccccc}\text{Sequence} & 5 && 10 && 17 && 26 && 37 \\<br />
\text{1st diff.} & & 5 && 7 && 9 && 11 \\<br />
\text{2nd diff.} & & & 2 & & 2 & & 2 \end{array}


    We see that the second differences are constant.
    This indicates that the generating function is of the second degree, a quadratic.
    . . That is, f(n) contains n^2.


    With that in mind, I looked the sequence again, and I saw:

    . . \begin{array}{c|ccc}<br />
n & & a_n \\ \hline<br />
1 & 5&=&2^2+1 \\<br />
2 & 10 &=&3^2+1 \\<br />
3 & 17 &=&4^2+1 \\<br />
4 & 26 &=&5^2+1 \\<br />
5 & 37 &=&6^2+1 \end{array}
    . . . \begin{array}{c|c}\vdots & \vdots \\<br />
n & \;\;(n+1)^2+1<br />
\end{array}


    Therefore: . f(n) \:=\:(n+1)^2+1 \:=\:n^2+2n+2

    . . which verifies u2_wa's result.

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