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Math Help - Consecutive integers...

  1. #1
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    Consecutive integers...

    The sum of c consecutive positive integers = c^2.
    Example, c=5: 3+4+5+6+7 = 25 ; 5^2 = 25
    (last integer = 7)

    If c = 9999, what is the last integer?

    What is the last integer in terms of c?
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  2. #2
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    Re: Consecutive integers...

    Code:
    c:=9999:
    for n from 1 to 10000 do
    if sum(k,k=n..n+c-1)=c^2 then
    print(n+c-1);
    break;
    end if;
    end do;
    for c=9999 , last integer =  14998

    in terms of c , last integer = \frac{3c-1}{2}
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  3. #3
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    Re: Consecutive integers...

    Good one Princeps.
    Accidentally got that while fooling around with consecutive numbers.
    Had you seen the "easy formula" last integer = (3c - 1)/2 before?
    Amazing that there is a solution for ALL odd c's.
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  4. #4
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    Re: Consecutive integers...

    a_i=\frac{c-1}{2}+i ~\text{ for }~ 1\leq i \leq c

    \displaystyle \sum_{i=1}^c \left(\frac{c-1}{2}+i\right)=c\cdot\frac{(c-1)}{2}+\displaystyle \sum_{i=1}^c i=\frac{c(c-1)}{2}+\frac{c(c+1)}{2}=c^2
    Last edited by princeps; April 11th 2012 at 06:47 AM.
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