October 29th, 2019, 08:31 AM | #1 |
Senior Member Joined: Aug 2018 From: România Posts: 110 Thanks: 7 | A sum
Hello all, Calculate $\displaystyle \sum_{k=n}^{k=1} k$. All the best, Integrator Last edited by Integrator; October 29th, 2019 at 08:58 AM. |
October 29th, 2019, 08:40 AM | #2 |
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 |
That doesn't mean anything. Did you forget some terms? The standard form is $\displaystyle \sum_{index~variable=starting~value}^{ending~value } (term~depending~on~index~variable)$ |
October 29th, 2019, 09:09 AM | #3 | |
Math Team Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 | Quote:
$\displaystyle \sum_{k=n}^{k=1} k = n + (n-1) + (n-2) + \, ... \, + 3 + 2 + 1$ Doesn't this end up with the same result as $\displaystyle \sum_{k=1}^n k$ ? If this is not what you meant, what is the relationship between $n$ and $1$ ? | |
October 29th, 2019, 09:09 AM | #4 |
Senior Member Joined: Aug 2018 From: România Posts: 110 Thanks: 7 | |
October 29th, 2019, 09:22 AM | #5 | |
Senior Member Joined: Aug 2018 From: România Posts: 110 Thanks: 7 | Quote:
All the best, Integrator | |
October 29th, 2019, 10:07 AM | #6 | |
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 | Quote:
I've never seen reverse indexing like that in mathematics before. My first instinct would just be to count backwards, like skeeter. My second instinct would be to interpret it like computer code for k = n to 1 step 1. If $k=1$, then sum the one term. if $k>1$, then no terms count and the sum is zero. | |
October 29th, 2019, 10:26 AM | #7 |
Math Team Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 | |
October 29th, 2019, 11:22 AM | #8 |
Math Team Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timey-wimey stuff. | No, that's fake news. Fake news. I mean, it could be something, but I don't think it really is. Fake means not right, not right, but fake. Fake news. That's all it is. Fake. How does my tie look? I tied it myself. It looks great. Really, I think it does. I like it. -Dan |
October 29th, 2019, 11:13 PM | #9 | |
Senior Member Joined: Aug 2018 From: România Posts: 110 Thanks: 7 | Quote:
It is true that the calculation of a sum is somewhat like the calculation of an integral? I say yes and then we can write that $\displaystyle \sum_{k=1}^{k=n} k=\frac{n+1}{2}+\int_1^nx dx$, which means that $\displaystyle \sum_{k=n}^{k=1} k=\frac{n+1}{2}+\int_n^1x dx=\frac{n+1}{2}+\frac{1-n^2}{2}=\frac{-n^2+n+2}{2}=-\frac{(n-2)(n+1)}{2}$. Is my reasoning correct? All the best, Integrator Last edited by skipjack; October 30th, 2019 at 12:46 AM. | |
October 30th, 2019, 12:59 AM | #10 |
Global Moderator Joined: Dec 2006 Posts: 21,107 Thanks: 2324 | |