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S Oct 18, 2022 at 20:14 history suggested Glorfindel CC BY-SA 4.0
typos corrected
Oct 18, 2022 at 19:30 review Suggested edits
S Oct 18, 2022 at 20:14
Sep 22, 2022 at 3:01 history tweeted twitter.com/StackMma/status/1572783069832290304
Sep 21, 2022 at 23:21 history became hot network question
Sep 21, 2022 at 19:38 vote accept Azzurro94
Sep 21, 2022 at 18:44 history edited Azzurro94 CC BY-SA 4.0
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Sep 21, 2022 at 18:38 history edited Azzurro94 CC BY-SA 4.0
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Sep 21, 2022 at 18:02 vote accept Azzurro94
Sep 21, 2022 at 18:38
Sep 21, 2022 at 17:59 answer added userrandrand timeline score: 2
Sep 21, 2022 at 17:18 comment added Azzurro94 @DanielHuber thanks. you are right but I need something like generic, symbolic one.
Sep 21, 2022 at 17:17 comment added Daniel Huber If you give k a definite values, you get ar[1]
Sep 21, 2022 at 17:17 answer added Bob Hanlon timeline score: 6
Sep 21, 2022 at 17:09 comment added userrandrand Not a general solution but for the example you gave you could use Sum[ar[j], { j, 1, k}] + Sum[- ar[j], { j, 2, k}] /. Sum[-a_, b_] :> - Sum[a, b] /. Sum[a_, {b1_, b2_, b3_}] :> Sum[a, {b1, 1, b3}] - Sum[a, {b1, 1, b2 - 1}]. That replacement rule uses Sum[ar[j], {j, 1, 0}]=0 in the first sum of your example.
Sep 21, 2022 at 16:30 comment added Bill I believe you will find that Mathematica does not automatically take Sums apart. I am guessing that is because of the many strange things that can happen when k turns out to be infinite and/or some of ar[j] turn out to be infinite or undefined or ... Think of the most malicious sums, things REALLY evil, and imagine what those borderline cases might do to a relatively mindless piece of software trying to follow a vast list of generic rules
Sep 21, 2022 at 15:21 history asked Azzurro94 CC BY-SA 4.0