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I am trying to get

   s[n_]:=Sum[1/2^k,{k,1,n}]

to show me the actual terms involved in the $n$th partial sum and not just the sum itself. I've tried Expand, Apart, etc. to no avail. Ideas?

Edit based on early comments: Is it necessary to use Table to accomplish this rather than some algebraic command applied to s[n] or some option of Sum?

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  • $\begingroup$ s[n_] := Row[Riffle[Table[1/2^k, {k, 1, n}], "+"]] then? $\endgroup$ – J. M.'s technical difficulties Nov 9 '12 at 16:01
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    $\begingroup$ Or s[n_Integer?Positive] := Array[1/2^# &, n, 1, Defer[Plus[##]] &] allowing you to copy/paste the result and evaluate it. $\endgroup$ – Sasha Nov 9 '12 at 16:04
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    $\begingroup$ @J.M. I am sure you are aware of the shortcut Row[Table[1/2^k, {k, 1, n}], "+"] to your code. $\endgroup$ – Sasha Nov 9 '12 at 16:05
  • $\begingroup$ @Sasha, not until you brought it up, actually; thanks! $\endgroup$ – J. M.'s technical difficulties Nov 9 '12 at 16:09
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    $\begingroup$ @J.M. Row[list,s] inserts s as a separator between successive elements. Damn! there is too much functionality to dig up $\endgroup$ – Dr. belisarius Nov 9 '12 at 18:37
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If working in the notebook the following

s[n_Integer?Positive] := Array[1/2^# &, n, 1, Defer[Plus[##]] &]

will produce an expression that can be copied, pasted and would evaluate as if the sum was explicitly typed in.

Alternatively you can build a visual resemblance of the sum using Row:

s[n_Integer?Positive] := Row[Table[1/2^k, {k, 1, n}], "+"]
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    $\begingroup$ That first one is so cool! ;-) $\endgroup$ – JohnD Nov 10 '12 at 19:00
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    $\begingroup$ @Sasha How would you handle this one: s[n_Integer?Positive] := Row[Table[(-1)^k/2^k, {k, 1, n}], "+"]. That is, how would you get the signs to alternate? $\endgroup$ – David Mar 15 '17 at 17:02
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    $\begingroup$ You would need to either use Riffle or Row[ Flatten[ {If[ Negative[#], "-", "+"], #}& /@ Table[ (-1)^k/2^k, {k, n}]]] $\endgroup$ – Sasha Mar 15 '17 at 17:12
  • $\begingroup$ f[n_Integer?Positive] := Sum[(-1)^k/2^k x^k, {k, 1, n}] /. {Plus :> Inactive[Plus], x -> 1} $\endgroup$ – Bob Hanlon Sep 10 '17 at 0:00

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