Timeline for Why is $\sum_{n=1}^{k}\frac{1}{n^2+n}=\frac{k}{k+1}$ [duplicate]
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| S Mar 11 at 23:52 | history | suggested | TShiong | CC BY-SA 4.0 |
better latexing
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| Mar 11 at 23:50 | review | Suggested edits | |||
| S Mar 11 at 23:52 | |||||
| Mar 11 at 21:15 | history | closed |
D S Martin R amWhy algebra-precalculus Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed. |
Duplicate of Prove the following equality using mathematical induction: $\sum\limits_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$ | |
| Feb 29 at 0:14 | vote | accept | Lucien Jaccon | ||
| Feb 26 at 13:50 | review | Close votes | |||
| Mar 11 at 21:15 | |||||
| Feb 26 at 13:32 | comment | added | Lucien Jaccon | Proof by induction doesn’t scratch that itch as much as derivation does. | |
| Feb 26 at 13:26 | answer | added | Eemil Wallin | timeline score: -1 | |
| Feb 26 at 13:18 | comment | added | Martin Sleziak | You could have a look, for example, at: What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ or Prove the following equality using mathematical induction: $\sum\limits_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$. | |
| Feb 26 at 13:15 | answer | added | Lucien Jaccon | timeline score: 3 | |
| Feb 26 at 13:03 | answer | added | Surb | timeline score: 3 | |
| Feb 26 at 13:00 | history | asked | Lucien Jaccon | CC BY-SA 4.0 |