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Math Help - cool matrix problem... Try this .. will be interesting

  1. #1
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    Post cool matrix problem... Try this .. will be interesting

    Matrix M1=[1];
    Matrix M2= 2 3
    ( 2*2 ) 4 5






    Matrix (3*3)M3 = 6 7 8
    9 10 11
    12 13 14



    Matrix(4*4) M4= 15 16 17 18
    19 20 21 22
    23 24 25 26
    27 28 29 30



    ..... etc for Matrix Mn


    Find the sum of elemnts of the main diagonal. and give the solution in terms of A+Bn+Cn^2...
    Last edited by kamaksh_ice; August 17th 2007 at 09:37 AM. Reason: I need the entire solution to this problem.
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  2. #2
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    Quote Originally Posted by kamaksh_ice View Post
    Matrix M1=[1];
    Matrix M2= 2 3
    ( 2*2 ) 4 5






    Matrix (3*3) M3 = 6 7 8
    9 10 11
    12 13 14



    Matrix(4*4) M4 = 15 16 17 18
    19 20 21 22
    23 24 25 26
    27 28 29 30



    ..... etc for Matrix Mn.






    Find the sum of elemnts of the main diagonal. and give the solution in terms of A+Bn+Cn^2...
    So you are asking for the general term of the series \{1, 8, 38, ... \} ?
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  3. #3
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    Hello, kamaksh_ice!

    M_1\:=\:\begin{pmatrix}1\end{pmatrix}\quad M_2\:=\:\begin{pmatrix}2 & 3 \\ 4 & 5\end{pmatrix} \quad M_3\:=\:\begin{pmatrix}6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14\end{pmatrix} \quad M_4 \:=\:\begin{pmatrix}15 & 16 & 17 & 18 \\ 19 & 20 & 21 & 22 \\ 23 & 24 & 25 & 26 \\ 27 & 28 & 29 & 30\end{pmatrix}
    ... and so on to matrix M_n.

    Find the sum of elements of the main diagonal.

    We have: . \begin{array}{cccc} 1 & = & 1 & [1] \\ 2 + 5 & = & 7 & [2]\\ 6 + 10 + 14 & = & 30 & [3]\\ 15 + 20 + 25 + 30 & = & 90 & [4]\\ 31 + 37 + 43 + 49 + 55 & = & 215 & [5]\end{array}


    Each is the sum of an arithmetic series.

    . . \begin{array}{cccc}\text{In [3]:} & a = 6 & d = 4 & n = 3 \\ \text{In [4]:} & a = 15 & d = 5  & n = 4 \\ \text{In [5]:} & a = 31 & d = 6 & n = 5\end{array}

    In the n^{th} matrix, the common difference is n+1, and there are n terms.


    We must determine the first terms of these series: 1,\,2,\,6,\,15,\,31,\,56\cdots

    After some algebra, we find that: . a \:=\:\frac{1}{6}(n+1)(2n^2-5n+6)


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

    So we have: . S \;=\;\frac{n}{2}\left[2\cdot\frac{1}{6}(n+1)(2n^2-5n+6) + (n-1)(n+1)\right]

    . . which simplifies to: . \boxed{S \;=\;\frac{n}{6}(n+1)(2n^2-2n+3)}

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