Timeline for Is there a closed form of $\sum_{i=1}^{n-k} {n-1-i\choose k-1}x^i$ in $x$?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Jun 18, 2021 at 6:41 | answer | added | 2734364041 | timeline score: 1 | |
| Jun 16, 2021 at 22:47 | comment | added | Max Alekseyev | Essentially you ask for truncation of the series $\sum_{m\geq k-1} \binom{m}{k-1} y^{m-k+1} = (1-y)^{-k}$. There seems to be no nice formula here. | |
| Jun 16, 2021 at 22:18 | review | Close votes | |||
| Jun 16, 2021 at 22:27 | |||||
| Jun 16, 2021 at 22:01 | comment | added | Benjamin L. Warren | In that question I was looking for ${n-2\choose k-1}1^a + {n-3\choose k-1}2^a...+{k-1\choose k-1}(n-k)^a$ but here I'm looking for ${n-2\choose k-1}x + {n-3\choose k-1}x^2...+{k-1\choose k-1}x^{n-k}$. | |
| Jun 16, 2021 at 21:58 | comment | added | Max Alekseyev | Then you need elaborate what is different and/or why you are not happy with answers given there. | |
| Jun 16, 2021 at 21:55 | comment | added | Benjamin L. Warren | This question is different in structure. | |
| Jun 16, 2021 at 21:54 | comment | added | Max Alekseyev | You have already asked this in Closed form of $ \sum_{i=1}^{n-k} {n-1-i\choose k-1}i^a + \sum_{i=1}^k {n-1-i\choose n-1-k}$ | |
| Jun 16, 2021 at 21:42 | comment | added | T. Amdeberhan | A closed formula does not seem feasible here. | |
| Jun 16, 2021 at 21:27 | history | asked | Benjamin L. Warren | CC BY-SA 4.0 |