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I want to expand the expression but I do not want it to evaluate.

Sum[(HoldForm[1/(2 # - 1) - 1/(2 # + 1)] &)[i], {i, 1, 6}]

However if you evaluate this you get: (Fraction-Fraction)+...+(Fraction-Fraction). What i want is to have this fraction evaluated without having to evaluate the subtraction inside the parenthesis or the addition outside the parenthesis. Could someone show me a simple way to do this?

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  • $\begingroup$ It is a good habit to hold on with an accept a day or two, let's don't discourage others. $\endgroup$
    – Kuba
    Commented Jan 17, 2015 at 16:24
  • $\begingroup$ @Kuba Okey, ill do that next time:) $\endgroup$
    – ALEXANDER
    Commented Jan 17, 2015 at 16:29

3 Answers 3

Reset to default
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HoldForm[# - #2] & @@@ Table[{1/(2 i - 1), 1/(2 i + 1)}, {i, 1, 6}] // Total

enter image description here

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  • $\begingroup$ I am still (till now) don't know how to format any of my answers that way you did it here. can you tell me how? Thanks. $\endgroup$
    – Basheer Algohi
    Commented Jan 18, 2015 at 1:30
  • $\begingroup$ @Algohi The result is a picture :) $\endgroup$
    – Kuba
    Commented Jan 18, 2015 at 8:03
  • $\begingroup$ Oh, I see. I should have known this. I can't copy it:-) $\endgroup$
    – Basheer Algohi
    Commented Jan 18, 2015 at 16:01
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One more subtle variation:

Sum[Defer[# - #2] & @@ (1/(2 i + {-1, 1})), {i, 6}]

(1/11 - 1/13) + (1/9 - 1/11) + (1/7 - 1/9) + (1/5 - 1/7) + (1/3 - 1/5) + (1 - 1/3)
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  • 1
    $\begingroup$ Always new thing to the scene. When the master speech, everyone has to listen. :) $\endgroup$
    – Basheer Algohi
    Commented Jan 17, 2015 at 20:49
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Another way of doing it (something similar to Kuba's great answer) is:

Sum[HoldForm[#1 - #2] &[1/(2 i - 1), 1/(2 i + 1)], {i, 1, 6}]

May be also something different:

Sum[(1/(2 i - 1) - 1/(2 i + 1) // Trace)[[-2]], {i, 1, 6}]
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    $\begingroup$ Nice! (8 to go) $\endgroup$
    – Kuba
    Commented Jan 17, 2015 at 22:08

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