Timeline for Sum of all products of k distinct integers in [1,n]
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2019 at 14:40 | vote | accept | Diger | ||
| Nov 21, 2019 at 14:34 | history | edited | Luis Ferroni | CC BY-SA 4.0 |
added 2 characters in body
|
| Nov 21, 2019 at 14:34 | comment | added | Luis Ferroni | Yes, you're right. I edit and fix the typo now. | |
| Nov 21, 2019 at 13:58 | comment | added | Diger | Hm, but $(x+1)\cdots(x+n)=\sum_{k=0}^n {n+1 \brack {k+1}} \, x^k$ and the coefficient of $x^k$ is the sum of all products of $n-k$ distinct integers in $\{1,...,n\}$. Therefore the coefficient of $x^{n-k}$ is the sum of all products of $k$ distinct integers which is ${n+1 \brack {n-k+1}}$. | |
| Nov 21, 2019 at 13:52 | comment | added | Luis Ferroni | I'm pretty sure it is what I said above.. | |
| Nov 21, 2019 at 11:48 | comment | added | Diger | Do you mean ${n+1 \brack {n-k+1}}$ ? | |
| Nov 21, 2019 at 0:30 | history | answered | Luis Ferroni | CC BY-SA 4.0 |