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Math Help - Chessboard Equation PLEASE HELP

  1. #1
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    Exclamation Chessboard Equation PLEASE HELP

    Hi,

    I am trying to work out a formula to tell you how many squares there are no a chessboard. I want to write the formula in a way that you could work it out for any size grid.

    The formula for a chessboard is x=8+7+6+5+4+3+2+1.

    Is there a function similar to Factorial but for addition not multiplication? This is so I could express the formula in the format of x=(n)! (replacing the factorial function for an appropriate one)?

    Any help would be greatly appreciated, as I am rusty on my mathematics and I'm trying to help my son with his maths project!

    PS: I have managed to write a program to work out the answer in the following, which loops until it has added the square of n until n is equal to 1. However, I don't think the maths teacher would be impressed with this, as I know she doesn't approve of computers doing the work for you!
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  2. #2
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    Hello, danielmegson!

    I am trying to work out a formula for how many squares there are on a chessboard.
    I want to write the formula in a way that you could work it out for any size grid.

    The formula for a chessboard is: . x \:=\: 1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2
    There is a formula for the sum of the first n squares: . S_n \;=\;\frac{n(n+1)(2n+1)}{6}

    For n = 8\!:\;\;S_8 \;=\;\frac{(8)(9)(17)}{6} \;=\;204



    The derivation of that formula takes quite a bit of algebra.
    . . If you're interested, I can show it to you.
    .
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  3. #3
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    FANTASTIC! SPOT ON! Thanks for the help! Any derivatives of this would be helpful as well, but if it's a lot of work to yourself please do not worry - you've been helpful enough!
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  4. #4
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    Hello again, Daniel!

    The derivation is not difficult, but it takes a number of steps.


    Consider the first few sums:

    . . \begin{array}{ccc}1^2 &=& 1 \\ 1^2+2^2 &=& 5 \\ 1^2+2^2+3^2 &=& 14 \\ 1^2+2^2+3^2+4^2 &=& 30 \\ 1^2+2^3+3^2+\hdots + 5^2 &=& 55 \\ 1^2+2^2+3^2 + \hdots + 6^2 &=&91 \\ 1^2+2^3+3^2+\hdots+7^2 &=& 140 \end{array}

    We want a function f(n) which generates these values.


    Take the difference of consecutive pairs of numbers.
    Then take the difference of the differences, and so on.

    \begin{array}{ccccccccccccc}<br />
\text{Sequence} & 1 && 5 && 14 && 30 && 55 && 91 \\<br />
\text{1st diff.} && 4 && 9 && 16 && 25 && 36 \\<br />
\text{2nd diff.} &&& 5 && 7 && 9 && 11 \\<br />
\text{3rd diff.} &&&&2 && 2 && 2 \end{array}


    Since the 3rd differences are constant,
    . . the function is of the 3rd degree . . . a cubic.

    The general cubic has the form: . f(n) \:=\:an^3 + bn^2 + cn + d


    Use the first four terms of our sequence:

    \begin{array}{ccccc}<br />
f(1)\:=\:1\!: & a + b + c + d &=& 1 & [1] \\<br />
f(2)\:=\:5\!: & 8a + 4b + 2c + d &=& 5 & [2] \\<br />
f(3) \:=\:14\!: & 27a + 9b + 3c + d &=& 14 & [3] \\<br />
f(4)\:=\:30\!: & 64a + 16b + 4c + d &=& 30 & [4] \end{array}


    \begin{array}{ccccc}<br />
\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 4 & [5] \\<br />
\text{Subtract [3] - [2]:} & 19a + 5b + c &=& 9 & [6] \\<br />
\text{Subtract [4] - [3]:} & 37a + 7b + c &=& 16 & [7] \end{array}


    \begin{array}{ccccc}<br />
\text{Subtract [6] - [5]:} & 12a + 2b &=& 5 & [8] \\<br />
\text{Subtract [7] - [6]:} & 18a + 2b &=& 7 &[9] \end{array}


    \begin{array}{ccccccc}<br />
\text{Subtract [9] - [8]:} & 6a \:=\:2 & \Rightarrow &\boxed{ a \:=\:\tfrac{1}{3}} \end{array}

    \begin{array}{cccccccc}\text{Substitute into [8]:} & 12\left(\tfrac{1}{3}\right) + 2b \;=\;5 & \Rightarrow &\boxed{ b \:=\:\tfrac{1}{2}} \end{array}

    \begin{array}{ccccccc}\text{Substitute into [5]:} & 7\left(\tfrac{1}{3}\right) + 3\left(\tfrac{1}{2}\right) + c \:=\:4 & \Rightarrow &\boxed{ c \:=\:\tfrac{1}{6}} \end{array}

    \begin{array}{cccccc}\text{Substitute into [1]:} & \tfrac{1}{3} + \tfrac{1}{2} + \tfrac{1}{6} + d \:=\:1 & \Rightarrow & \boxed{d \:=\:0} \end{array}


    Therefore: . f(n) \;=\;\tfrac{1}{3}n^3 + \tfrac{1}{2}n^2 + \tfrac{1}{6}n \quad\Rightarrow\quad\boxed{ f(n) \;=\;\frac{n(n+1)(2n+1)}{6}}

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