0
$\begingroup$

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 $

$\endgroup$
7
  • $\begingroup$ Yes you are. Explictly compute all summations for $m=3$. $\endgroup$
    – user65203
    Nov 2 '16 at 22:22
  • $\begingroup$ $\sum_{k=1}^m k = 1 + 2 + \cdots + m \ne k m$. $\endgroup$
    – dxiv
    Nov 2 '16 at 22:25
  • $\begingroup$ @dxiv oh dang.. why do i think its a constant.. good looks $\endgroup$
    – John
    Nov 2 '16 at 22:28
  • $\begingroup$ @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. $\endgroup$
    – epi163sqrt
    Nov 3 '16 at 11:14
  • 1
    $\begingroup$ @MarkusScheuer Sorry, won't do it again. I continue to learn. $\endgroup$ Nov 3 '16 at 11:46
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.