A sum

skipjack

Forum Staff
Dec 2006
21,478
2,470
Some say that \(\displaystyle \sum_{k=1}^{k=n} k+\sum_{k=n}^{k=1} k =n+1\)
For $n \geqslant 1$, \(\displaystyle \sum_{k=1}^{k=n} k = \frac12n(n + 1)\).

For $n \leqslant 1$, \(\displaystyle \sum_{k=n}^{k=1} k = -\frac12(n - 2)(n + 1)\).

\(\displaystyle \frac12n(n + 1) - \frac12(n - 2)(n + 1) = n + 1\)
 
Aug 2018
137
7
România
For $n \geqslant 1$, \(\displaystyle \sum_{k=1}^{k=n} k = \frac12n(n + 1)\).

For $n \leqslant 1$, \(\displaystyle \sum_{k=n}^{k=1} k = -\frac12(n - 2)(n + 1)\).

\(\displaystyle \frac12n(n + 1) - \frac12(n - 2)(n + 1) = n + 1\)
Hello,

I do not understand!From the "WolframAlpha" read for \(\displaystyle n>0\):

https://www.wolframalpha.com/input/?i=+sum_%28k%3Dn%29%5E1+k.

How can you explain this result given by "WolframAlpha"?!?!
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And about https://www.wolframalpha.com/input/?i=+sum_%28k%3Dn%29%5E1+k%5E2 and https://www.wolframalpha.com/input/?i=sum_%28k%3D1%29%5En+k%5E2+%2B+sum_%28k%3Dn%29%5E1+k%5E2%3Dn%5E2%2B1 , what can you say?Thank you very much!

All the best,

Integrator
 
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