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Math Help - Division proof

  1. #1
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    Division proof

    Prove that (1+2+\cdots+9)|1^5+2^5+\cdots+9^5.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    here is how you do it....I will do a general proof...ok you want to prove that \sum_{i=0}^n{i}|\sum_{i=0}^n{i^5} to do this all you need to do is utilize summation formulas and see that \sum_{i=0}^n{i}=\frac{n(n+1)}{2} and that \sum_{i=0}^m{i^5}=\frac{n^2(n+1)^2(2n^2+2n-1)}{12} then we need to prove that \frac{\sum_{i=0}^n{i^5}}{\sum_{i=0}^n{i}} we do this by using substitution..the former thing can be rewritten as \frac{\frac{n^2(n+1)^2(2n^2+2n-1)}{12}}{\frac{n(n+1)}{2}} which simplifies to \frac{n(n+1)(2n^2+2n-1)}{6} now imputing 9 we get 2685...therfore we have proved your task
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  3. #3
    Super Member PaulRS's Avatar
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    It's not difficult to prove it for any odd exponent <br />
{2a + 1}

    We want to show that <br />
\left. {\sum\limits_{j = 1}^n j } \right|\sum\limits_{j = 1}^n {j^{2a + 1} }

    Case  <br />
n = 2 \cdot k<br />

    <br />
\sum\limits_{j = 1}^{2k} {j^{2a + 1} }  = \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 0}^{k - 1} {\left( {2 \cdot k - j} \right)^{2a + 1} }  \equiv \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 0}^{k - 1} {\left( { - j} \right)^{2a + 1} }=k^{2a+1} \left( {\bmod .k} \right)

    <br />
 \Rightarrow \left. k \right|\sum\limits_{j = 1}^{2k} {j^{2a+1} } <br />

    <br />
\sum\limits_{j = 1}^{2k} {j^{2a + 1} }  = \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 1}^k {\left( {2 \cdot k + 1 - j} \right)^{2a + 1} }  \equiv \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 1}^k {\left( { - j} \right)^{2a + 1} } \left( {\bmod .2k + 1} \right)<br />

    <br />
 \Rightarrow \left. (2k+1) \right|\sum\limits_{j = 1}^{2k} {j^{2a+1} } <br />

    Thus: <br />
\left. {\sum\limits_{j = 1}^{2k} j } \right|\sum\limits_{j = 1}^{2k} {j^{2a + 1} } since <br />
{k \cdot \left( {2k + 1} \right) = \sum\limits_{j = 1}^{2k} j } and <br />
\left( {2k + 1;k} \right) = 1



    Case:  <br />
n = 2 \cdot k + 1<br />

    <br />
\sum\limits_{j = 1}^{2k + 1} {j^{2a + 1} }  = \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 0}^k {\left( {2 \cdot k + 1 - j} \right)^{2a + 1} }  \equiv \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 0}^k {\left( { - j} \right)^{2a + 1} } \left( {\bmod .2k + 1} \right)<br />

    <br />
 \Rightarrow \left. (2k+1) \right|\sum\limits_{j = 1}^{2k+1} {j^{2a+1} } <br />


     <br />
\sum\limits_{j = 1}^{2k + 1} {j^{2a + 1} }  = \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 1}^{k + 1} {\left( {2 \cdot \left( {k + 1} \right) - j} \right)^{2a + 1} } <br /> <br />
and

     <br /> <br />
\sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 1}^{k + 1} {\left( {2 \cdot \left( {k + 1} \right) - j} \right)^{2a + 1} }  \equiv \sum\limits_{j = 1}^k {j^{2a + 1} }  + \sum\limits_{j = 1}^{k + 1} {\left( { - j} \right)^{2a + 1} }  \equiv 0\left( {\bmod .k + 1} \right)<br />

    <br />
 \Rightarrow \left. (k+1) \right|\sum\limits_{j = 1}^{2k+1} {j^{2a+1} } <br />

    Thus: <br />
\left. {\sum\limits_{j = 1}^{2k+1} j } \right|\sum\limits_{j = 1}^{2k+1} {j^{2a + 1} } since <br />
{(k+1) \cdot \left( {2k + 1} \right) = \sum\limits_{j = 1}^{2k+1} j } and <br />
\left( {2k + 1;k+1} \right) = 1

    Therefore for all <br />
n,a \in \mathbb{Z}^ +  <br />
we have <br />
\left. {\sum\limits_{j = 1}^n j } \right|\sum\limits_{j = 1}^n {j^{2a + 1} } <br />
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