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Math Help - series

  1. #1
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    series

    the natural numbers are grouped as follows:
    (1) , ( 2 , 3 ) , ( 4 , 5 , 6 ) , ( 7 , 8 , 9 , 10 ) , ...

    find an expression for the first term of the nth group and prove that the sum of the numbers in the nth group is 1/2 n (n^2+1).

    uh...I have no idea where to even start.
    thank you in advance!
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by colloquial View Post
    the natural numbers are grouped as follows:
    (1) , ( 2 , 3 ) , ( 4 , 5 , 6 ) , ( 7 , 8 , 9 , 10 ) , ...

    find an expression for the first term of the nth group and prove that the sum of the numbers in the nth group is 1/2 n (n^2+1).

    uh...I have no idea where to even start.
    thank you in advance!
    Here is a pattern i found. I hope it helps.
    Attached Files Attached Files
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  3. #3
    MHF Contributor

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    The nth grouping will be \left( {\frac{{n^2  - n + 2}}{2}, \cdots ,\frac{{n^2  + n}}{2}} \right).
    Now apply the Gauss formula for the sum of the first n positive integers.
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  4. #4
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    Hello, colloquial!

    The natural numbers are grouped as follows:
    \{1\},\;\{2,3\},\;\{4,5,6\},\;\{7,8,9,10\},\;...

    (a) Find an expression for the first term of the n^{th}group.

    (b) Prove that the sum of the numbers in the n^{th} group is: . \frac{1}{2}n(n^2+1)

    Note the last number of each group: . 1,\,3,\,6,\,10,\,\cdots
    . . These are the triangular numbers: . T_n \:=\:\frac{n(n+1)}{2}
    So each group begins with the preceding triangular number, plus 1.

    (a) The first term of the n^{th} group is: . a_n \:=\:\frac{n(n-1)}{2}+1\;=\;{\color{blue}\frac{n^2-n+2}{2}}


    (b) Each group is an arithmetic series with first term, a_1 \:=\:\frac{n^2-n+2}{2},
    . . common difference, d = 1, and n terms.

    The sum of an arithmetic series is: . S_n \;=\;\frac{n}{2}\left[2a_1 + (n-1)d\right]

    Therefore: . S_n \;=\;\frac{n}{2}\left[2\!\cdot\!\frac{n^2-n+2}{2} + (n-1)\!\cdot\!1\right] \;=\;{\color{blue}\frac{1}{2}n(n^2+1)}

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