Algebra Pre-Algebra and Basic Algebra Math Forum

 November 1st, 2019, 10:52 PM #21 Senior Member   Joined: Aug 2018 From: România Posts: 110 Thanks: 7 Hello all, From the "WolframAlpha" read: https://www.wolframalpha.com/input/?...E2%3Dn%5E2%2B1. What do you think about this shocking result! All the best, Integrator November 2nd, 2019, 01:20 AM   #22
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Quote:
 Originally Posted by Integrator 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 $n \leqslant 1$, $\displaystyle \sum_{k=n}^{k=1} k = -\frac12(n - 2)(n + 1)$.

$\displaystyle \frac12n(n + 1) - \frac12(n - 2)(n + 1) = n + 1$ November 2nd, 2019, 07:38 AM   #23
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Quote:
 Originally Posted by skipjack 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/?...k%3Dn%29%5E1+k.

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

All the best,

Integrator

Last edited by Integrator; November 2nd, 2019 at 07:51 AM. Tags sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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