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I have to calculate the following summation, and have no way of knowing if I am right and feel like I am doing this wrong.. If so can you help me do it properly?

$ \sum_{j=1}^{m} \sum_{k=1}^{m} jk $

  1. $ \sum_{k=1}^{m} jk => j * \sum_{k=1}^{m} k => j(km)$

  2. $ \sum_{j=1}^{m} j(km) => km \sum_{j=1}^{m} j => km(jm) = kjm^2 \square $

hamam_Abdallah
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John
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  • Yes you are. Explictly compute all summations for $m=3$. –  Nov 02 '16 at 22:22
  • $\sum_{k=1}^m k = 1 + 2 + \cdots + m \ne k m$. – dxiv Nov 02 '16 at 22:25
  • @dxiv oh dang.. why do i think its a constant.. good looks – John Nov 02 '16 at 22:28
  • @AbdallahHammam: Please note, it is not an appropriate way to give a hint by modifying the question. We should not change the content/meaning of the question. If another user looks at this question it is no longer clear what OP was asking and you prevent this way other people to effectively support OP. You should correct the question and restore the original version. It is much better to add a comment or answer instead. – Markus Scheuer Nov 03 '16 at 11:14
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    @MarkusScheuer Sorry, won't do it again. I continue to learn. – hamam_Abdallah Nov 03 '16 at 11:46
  • @AbdallahHammam: Everything's fine! Many thanks for your quick correction. – Markus Scheuer Nov 03 '16 at 11:50
  • $j$ and $k$ are dummy variables, John – they can't possibly appear in the answer. Only $m$ can. – Gerry Myerson Nov 03 '16 at 11:52

2 Answers2

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The calculation should be revised. Recall the sum of the first $m$ natural numbers is \begin{align*} \sum_{k=1}^mk=1+2+3+\cdots+m=\frac{m(m+1)}{2}\tag{1} \end{align*}

We obtain \begin{align*} \sum_{j=1}^m\sum_{k=1}^mjk&=\sum_{j=1}^mj\left(\sum_{k=1}^mk\right)\tag{2}\\ &=\left(\sum_{j=1}^mj\right)\left(\sum_{k=1}^mk\right)\tag{3}\\ &=\left(\sum_{j=1}^mj\right)^2\tag{4}\\ &=\left(\frac{m(m+1)}{2}\right)^2\tag{5}\\ &=\frac{m^2(m+1)^2}{4}\tag{6}\\ \end{align*}

Comment:

  • In (2) we factor out $j$ from the inner sum.

  • In (3) we use the distributive property of $*$ over $+$ and obtain a representation as product of two series.

  • In (4) we observe that both series are the same and so we can write the product as square of the series.

  • In (5) we use the formula (1).

  • In (6) we do a final simplification.

Markus Scheuer
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I can immediately say that your answer is wrong because it uses the variables of summation.

The result should be a function of m only.

marty cohen
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