Timeline for Specific partial sum of even/odd binomial coefficients
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2019 at 19:49 | comment | added | user64494 | Thank you. How is A000346 useful to the question under consideration? TIA. | |
| Feb 28, 2019 at 19:34 | comment | added | Robert Israel | $S_g = (2 g+1) A000346(2 g-1)$. | |
| Feb 28, 2019 at 17:57 | comment | added | user64494 | This is $ a(n) = 2^{2n+1} $- $ {2n+1}\choose {n+1}$. How is this sequence related to the sum under consideration? | |
| Feb 28, 2019 at 15:36 | comment | added | Robert Israel | See also OEIS sequence A000346 | |
| Feb 28, 2019 at 14:26 | vote | accept | piogor | ||
| Feb 28, 2019 at 13:55 | comment | added | piogor | Thanks to your answer, I managed to find the proof. Please feel free, to include it in your post. Using $\binom{4g+1}{2k-1}=\binom{4g}{2k-1}+\binom{4g}{2k-2}$: $\sum_{k=1}^g \binom{4g+1}{2k-1}=\sum_{k=1}^g \binom{4g}{2k-1}+\sum_{k=0}^{g-1} \binom{4g}{2k}=\sum_{k=0}^{2g-1} \binom{4g}{k}=\dfrac{1}{2}(2^{4g}-\binom{4g}{2g})$ | |
| Feb 28, 2019 at 13:27 | history | answered | Robert Israel | CC BY-SA 4.0 |