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edited Apr 13, 2017 at 12:55

As a workaround for this incorrect simplification, you could use FindSequenceFunction on a sample of concrete results:

Clear[m,n,k];
Simplify[FindSequenceFunction[
   Table[
    Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}], {m, 2, 12, 2}],
    n] /. n -> m/2]

(* ==> 2^(-4 + m) (-2 + m) *)

This is similar to the answer by @Coolwater here here.

As a workaround for this incorrect simplification, you could use FindSequenceFunction on a sample of concrete results:

Clear[m,n,k];
Simplify[FindSequenceFunction[
   Table[
    Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}], {m, 2, 12, 2}],
    n] /. n -> m/2]

(* ==> 2^(-4 + m) (-2 + m) *)

This is similar to the answer by @Coolwater here.

As a workaround for this incorrect simplification, you could use FindSequenceFunction on a sample of concrete results:

Clear[m,n,k];
Simplify[FindSequenceFunction[
   Table[
    Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}], {m, 2, 12, 2}],
    n] /. n -> m/2]

(* ==> 2^(-4 + m) (-2 + m) *)

This is similar to the answer by @Coolwater here.

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created Feb 9, 2016 at 0:37

Jens

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As a workaround for this incorrect simplification, you could use FindSequenceFunction on a sample of concrete results:

Clear[m,n,k];
Simplify[FindSequenceFunction[
   Table[
    Sum[Binomial[m - 2, k - 1] (k - 1), {k, 2, m, 2}], {m, 2, 12, 2}],
    n] /. n -> m/2]

(* ==> 2^(-4 + m) (-2 + m) *)

This is similar to the answer by @Coolwater here.