Say I want to create the list
$\{ \sum_{j=1}^{5} 2j, \sum_{j=1}^{6} \frac{5j}{2} \}$
I guessed that this can be done using the code
Sum[j/{1, 2}*{2, 5}, {j, 1, {5, 6}}]
However what mathematica gives is
$\Big{\{} \{ \sum_{j=1}^{5} 2j, \sum_{j=1}^{5} \frac{5j}{2} \}, \{ \sum_{j=1}^{6} 2j, \sum_{j=1}^{6} \frac{5j}{2} \} \Big{\}}$
I guess I could just somehow pick out the right element of the result, but is there a more efficient way to achieve this?

EDIT: Eventually I want to apply this to a case, where I have something like

  • 1
    $\begingroup$ Maybe: MapThread[Sum[# j, {j, 1, #2}] &, {{2, 5/2}, {5, 6}}] $\endgroup$ – Coolwater Jun 4 '19 at 15:03
  • $\begingroup$ How can I apply this to an arbitrary list and an arbitrary function of that list, e.g. something like Sum[f[list,j],{j,1,list}]? $\endgroup$ – lomby Jun 4 '19 at 16:14

Your question is not very well-defined. It seems that you have a list of different functions, f1, f2, f3, etc., and you want to sum each of those functions to different maximum values. Is that correct? If so:

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sum[a_, b_, c_] := Sum[a j, {j, b, c}]
SetAttributes[sum, Listable]


sum[{2, 5}/{1, 2}, 1, {5, 6}]

{30, 105/2}

sum[{a, b}, 1, {n, m}]

{1/2 a n (1 + n), 1/2 b m (1 + m)}

sum[{foo[j], bar[j]}, q, {n, m}] // TeXForm

$\left\{\sum _{j=q}^n j \text{foo}(j),\sum _{j=q}^m j \text{bar}(j)\right\}$

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