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Math Help - counting number

  1. #1
    Junior Member
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    counting number

    i have the question
    The counting numbers are written in the pattern at the righ. Find the midle number of the 40th row.

    1
    2 3 4
    5 6 7 8 9
    10 11 12 13 14 15 16 [/LEFT]
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  2. #2
    Senior Member nikhil's Avatar
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    solution is 1561.
    simple method is to make general term of the middle element of each row using A.P then finding the 40th element
    general term will be n(n-1)+1
    n=1 gives 1
    n=2 gives 3
    n=3 gives 7
    these all are middle term of each row so on putting 40 middle number of 40th row will be obtained
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  3. #3
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    Hello, rai2003!

    \text{The counting numbers are written in this pattern:}

    . . . . . . \begin{array}{ccccccc}<br />
&&&1 \\ && 2 & 3 & 4 \\ & 5 & 6 & 7 & 8 & 9 \\<br />
10 & 11 & 12 & 13 & 14 & 15& 16 \\<br />
& & & \vdots\end{array}


    \text{Find the middle number of the }40^{th}\text{ row.}


    The middle numbers form this sequence: . 1,\:3,\:7,\:13,\:21,\:\hdots


    Take the differences of consecutive terms,
    . . then the differences of the differences and so on.

    . . \begin{array}{cccccccccccccc}<br />
\text{Sequence} & 1 && 3 && 7 && 13 && 21 && \hdots \\<br />
\text{1st difference} && 2 && 4 && 6 && 8 && \hdots\\<br />
\text{2nd difference} &&& 2 && 2 && 2 && \hdots \end{array}


    Since the second differences are constant,
    . . the generating function is of the second degree . . . a quadratic.

    The general quadratic function is: . f(n) \:=\:an^2 + bn + c


    We know the first term terms of the sequence.

    . . \begin{array}{cccccccccc}<br />
f(1) = 1: & a + b + c &=& 1 & [1] \\<br />
f(2) = 3: & 4a + 2b + c &=& 3 & [2] \\<br />
f(3) = 7: & 9a + 3b + c &=& 7 & [3] \end{array}


    \begin{array}{cccccccc}<br />
\text{Subtract [2] - [1]:} & 3a + b &=& 2 & [4] \\<br />
\text{Subtract [3] - [2]:} & 5a + b &=& 4 & [5] \end{array}


    \begin{array}{ccccccc}<br />
\text{Subtract [5] - [4]:} & 2a \:=\: 2 & \Rightarrow & a \:=\:1 \end{array}

    Substitute into [4]: . 3(1) + b \:=\:2 \quad\Rightarrow\quad b \:=\:\text{-}1

    Substitute into [1]: . 1 + (\text{-}1) + c \:=\:1 \quad\Rightarrow\quad c \:=\:1


    Hence, the generatng function is: . f(n) \;=\;n^2-n+1


    Therefore: . f(40) \;=\;40^2 - 40 + 1 \;=\;\boxed{1561}
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  4. #4
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    thanks
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