Timeline for Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2013 at 23:01 | comment | added | Steven Stadnicki | Gorgeous! Now all we need is a combinatorial (counting) argument... :-) | |
| Oct 6, 2013 at 11:12 | history | edited | Mark Wildon | CC BY-SA 3.0 |
r should have been between 1 and m, not 1 and m-1
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| Oct 6, 2013 at 4:15 | vote | accept | mathlove | ||
| Oct 5, 2013 at 23:49 | history | undeleted | Mark Wildon | ||
| Oct 5, 2013 at 23:47 | history | deleted | Mark Wildon | via Vote | |
| Oct 5, 2013 at 23:39 | history | answered | Mark Wildon | CC BY-SA 3.0 |