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added 30 characters in body

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edited Feb 11, 2016 at 18:18

This is a bug in SumSum.

In addition to the nice workarounds proposed by others, the substitution m = 2p$m = 2p$ also leads to a correct answer for this example.

In[1]:= Sum[ Binomial[2 p - 2, k - 1] (k - 1), {k, 2, 2 p - 2, 2}] /. {p -> m/2} // Simplify

Out[1]= 2^(-4 + m) (-2 + m)

In[1]:= Sum[
   Binomial[2 p - 2, k - 1] (k - 1), {k, 2, 2 p - 2, 2}] /. {p -> 
    m/2} // Simplify


Out[1]= 2^(-4 + m) (-2 + m)

Sorry for the inconvenience caused by this problem.

This is a bug in Sum.

In addition to the nice workarounds proposed by others, the substitution m = 2p also leads to a correct answer for this example.

In[1]:= Sum[ Binomial[2 p - 2, k - 1] (k - 1), {k, 2, 2 p - 2, 2}] /. {p -> m/2} // Simplify

Out[1]= 2^(-4 + m) (-2 + m)

Sorry for the inconvenience caused by this problem.

This is a bug in Sum.

In addition to the nice workarounds proposed by others, the substitution $m = 2p$ also leads to a correct answer for this example.

In[1]:= Sum[
   Binomial[2 p - 2, k - 1] (k - 1), {k, 2, 2 p - 2, 2}] /. {p -> 
    m/2} // Simplify


Out[1]= 2^(-4 + m) (-2 + m)

Sorry for the inconvenience caused by this problem.

Source Link

created Feb 9, 2016 at 17:28

This is a bug in Sum.

In addition to the nice workarounds proposed by others, the substitution m = 2p also leads to a correct answer for this example.

In[1]:= Sum[ Binomial[2 p - 2, k - 1] (k - 1), {k, 2, 2 p - 2, 2}] /. {p -> m/2} // Simplify

Out[1]= 2^(-4 + m) (-2 + m)

Sorry for the inconvenience caused by this problem.