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Math Help - How to form the correct equation

  1. #1
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    Question How to form the correct equation

    I suspect I am in the correct forum. However, due to partial or complete ignorance on my part, of disciplines above and including algebra, I may not be.
    So in an effort save some time for this particular situation, I am presenting this here, before I embark on the learning curve,...if that is even possible.

    Trying to define the equation I would use to find the value of a given number, using the following logic:

    Given Number: 2 its value is: 1 determined via 1
    Given Number: 3 its value is: 3 determined via 1+2
    Given Number: 4 its value is: 6 determined via 1+2+3
    Given Number: 5 its value is: 10 determined via 1+2+3+4
    ... to infinity

    This feels obvious,...but I just can't seem to see it.
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  2. #2
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    Try \displaystyle t_n= 1+2+3+\cdots +(n-1)+ n = \frac{n(n+1)}{2}
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Try \displaystyle t_n= 1+2+3+\cdots +(n-1)+ n = \frac{n(n+1)}{2}
    \displaystyle \frac{n(n+1)}{2}{-n} produces the correct result.

    However, (and again please excuse my ignorance), it is unclear to me how you arrived at: \displaystyle \frac{n(n+1)}{2}

    from: \displaystyle 1+2+3+\cdots +(n-1)+ n
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  4. #4
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    Quote Originally Posted by spfaff View Post
    [tex]However, (and again please excuse my ignorance), it is unclear to me how you arrived at: \displaystyle \frac{n(n+1)}{2}
    from: \displaystyle 1+2+3+\cdots +(n-1)+ n
    That is just one of the best known formulas in all mathematics.
    It in easily proven using mathematical induction.
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  5. #5
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    There is a story that when Gause was 8 years old, just to keep the students busy, his teacher assigned them the problem of adding all integer from 1 to 100. Gause immediately wrote 5050 on his slate, turned it over, and sat there. Presumably, he apparently recognized that if he reversed the sum he would have
    \begin{array}{ccccc}1+ & 2+ & + 3& +\cdot\cdot\cdot &+ 100 \\ 100+ & 99+ & 98+ \cdot\cdot\cdot&+ 1\\ 101 & + 101& + 101& \cdot\cdot\cdot & 101\end{array}
    That is, each pair adds to 100 and there are 100 such pairs: 101(100)= 10100 and since we have added twice each sum is 10100/2= 5050.

    Similarly, you can continue to "n" rather than 100 to get
    \begin{array}{ccccc}1+ & 2+ & + 3& +\cdot\cdot\cdot &+ n \\ n+ & n-1+ & n-2+ \cdot\cdot\cdot&+ 1\\ n+1 & + n+1&+ n+ 1&+ n+1& \cdot\cdot\cdot & n+1\end{array}
    so there are n pairs, each summing to n+1: the total of the pairs is n(n+1) and the sum of each is
    \frac{n(n+2)}{2}
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