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Here's my approach using replacement rule /. Sum[r^p_. a__, b_] :> r^p Sum[Times[a], b]. I have a hunch that there's a built in function that could accomplish this but I can't seem to find it. I've also made some modifications to your code: 1) Changing the first function to β to avoid clashing with the subscripted symbol α. 2) Changing the upper limit ...


I think you may find value in this rule: sequentialQ[x_List] := x == Range[x[[1]], x[[-1]]] rule = HoldPattern[Plus[s : F_[_, j___] ..]] /; sequentialQ @ {s}[[All, 1]] :> (HoldForm[Sum[F[i, j], {i, ##}]] & @@ {Min@#, Max@#} &[{s}[[All, 1]]]); Test: F[1.0, q, r] + F[1.5, q, r] + F[2, q, r] + F[2.7, q, r] + F[3.0, q, r] /. rule


In the question, there is no summation over $\nu$, as there is in other answers. In case $\nu$ is not specified, as the OP explicitly states, and meant to be an index, here is a way to calculate the desired tensor without using Table and Sum. SymbolicTensors`ArrayContract[ TensorContract[ a \[TensorProduct] b \[TensorProduct] c \[TensorProduct] ...


Assuming $\mu$ and $\nu$ also run from 1 to 4 (which they have to, otherwise your expression doesn't make sense), you can simply take a cue from this Q&A and write TensorContract[ TensorProduct[LeviCivitaTensor[4], a, b, c], {{2, 5}, {3, 6}, {4, 7}} ] // Normal { -a4 b3 c2 + a3 b4 c2 + a4 b2 c3 - a2 b4 c3 - a3 b2 c4 + a2 b3 c4, a4 b3 c1 - a3 ...


You need to construct a "table" that acts like a vector a = {a1, a2, a3, a4}; b = {b1, b2, b3, b4}; c = {c1, c2, c3, c4}; Table[ Sum[LeviCivitaTensor[4][[mu, nu, alpha, beta]] a[[nu]] b[[alpha]] c[[beta]], {alpha, 1, 4}, {beta, 1, 4}, {nu, 1, 4}], {mu, 1, 4}]

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