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Math Help - Calculate !

  1. #1
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    Arrow Calculate !

    Good afternoon everyone !

    I'm supposed to calculate this sum !

    Calculate !-dsc_0486-1.jpg

    I know that in the end, i'm supposed to get something like : (n(n+1)(4n-1))/6 or if you dev. this expression : 5n/6...

    Could anybody give me the step-by-step resolution of this exercice ? I have really no idea about what i'm supposed to do with it. I tried to transform it into a double sum, but that's all...


    Last edited by Kazey; October 9th 2013 at 10:32 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Calculate !

    Quote Originally Posted by Kazey View Post
    Good afternoon everyone !

    I'm supposed to calculate this sum !

    Click image for larger version. 

Name:	DSC_0486-1.jpg 
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Size:	300.1 KB 
ID:	29427

    I know that in the end, i'm supposed to get something like : (n(n+1)(4n-1))/6 or if you dev. this expression : 5n/6...

    Could anybody give me the step-by-step resolution of this exercice ? I have really no idea about what i'm supposed to do with it. I tried to transform it into a double sum, but that's all...


    You aren't sure if n(n + 1)(4n - 1)/6 is right?? Well, it's correct.

    I'd do it inductively. Your base case is n = 1 and it works just fine.

    So assume that \sum_{1 \leq i, j \leq k} max(i, j) = \frac{k(k + 1)(4k - 1)}{6} for some value of k.

    Then you need to show that
    \sum_{1 \leq i, j \leq k + 1} max(i, j) = \frac{(k + 1)((k + 1) + 1)(4(k + 1) - 1)}{6}

    So. I'll give you an outline, let's see what you can do with it. You need to show that
    \sum_{1 \leq i, j \leq k + 1} max(i, j) = \sum_{1 \leq i, j \leq k} max(i, j) + \text{Other terms}

    How many terms do you need to add to the k case on the RHS to make it equal to the k + 1 case?

    -Dan
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