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Math Help - Summing Series -2 questions.

  1. #1
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    Red face Summing Series -2 questions.

    Hey, I've got 2 unsolved problems that need help!

    1. Find the sum of the multiples of 7 which are less than 10000.
    My answer is 714264285 but the book says 7142142?

    2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n.
    A step by step guide would be really appreciated for this one.

    Thanks in advance!
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Nyoxis View Post
    Hey, I've got 2 unsolved problems that need help!

    1. Find the sum of the multiples of 7 which are less than 10000.
    My answer is 714264285 but the book says 7142142?
    There are 1428 multiples of 7 between 1 and 10000, because \frac{10000}{7} \approx 1428.571428571429.
    Therefore, you can write the sum this way :

    S=7 \cdot 1+7 \cdot 2+7 \cdot 3+\dots+7 \cdot 1427+7 \cdot 1428

    S=7 \left(1+2+\dots+1427+1428\right)

    What's in brackets is the sum of the 1428 first integers.

    We know that 1+\dots+n=\frac{n(n+1)}{2}

    So here, 1+2+\dots+1428=\frac{1428 \cdot 1429}{2}=\boxed{1020306}

    Hence S=7 \cdot 1020306=\boxed{7142142}



    How did you find 714264285 ?
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  3. #3
    Moo
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    Quote Originally Posted by Nyoxis View Post
    2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n.
    A step by step guide would be really appreciated for this one.
    Write it this way :

    S=\sum_{k=0}^{n-1} (k+1)(n-k)

    S=\sum_{k=0}^{n-1} \left[nk+n-k-k^2\right]=\sum_{k=0}^{n-1} (n-1)k+\sum_{k=0}^{n-1} n-\sum_{k=0}^{n-1} k^2

    S=(n-1) \sum_{k=0}^{n-1} k+n \sum_{k=0}^{n-1} 1-\sum_{k=0}^{n-1} k^2



    We know that \sum_{k=0}^n k=\sum_{k=1}^n k=\frac{n(n+1)}{2}. Be careful, here it's n-1, not n.

    We also know that \sum_{k=0}^n k^2=\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}. Be careful here too



    There may be a simplier way, but I just can't see it..
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