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 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 $\displaystyle \sum_{index~variable=starting~value}^{ending~value } (term~depending~on~index~variable)$ Thanks from idontknow
October 29th, 2019, 09:09 AM   #3
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Quote:
 Originally Posted by Integrator Hello all, Calculate $\displaystyle \sum_{k=n}^{k=1} k$. All the best, Integrator
so, assuming $n > 1$, is this what you mean?

$\displaystyle \sum_{k=n}^{k=1} k = n + (n-1) + (n-2) + \, ... \, + 3 + 2 + 1$

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
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 Originally Posted by DarnItJimImAnEngineer That doesn't mean anything. Did you forget some terms? The standard form is $\displaystyle \sum_{index~variable=starting~value}^{ending~value } (term~depending~on~index~variable)$
I corrected! Thank you very much!

All the best,

Integrator

October 29th, 2019, 09:22 AM   #5
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 Originally Posted by skeeter so, assuming $n > 1$, is this what you mean? $\displaystyle \sum_{k=n}^{k=1} k = n + (n-1) + (n-2) + \, ... \, + 3 + 2 + 1$ 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 $1$ ?
Some say that $\displaystyle \sum_{k=n}^{k=1} k$ it's not equal with $\displaystyle \sum_{k=1}^{k=n} k$ where $\displaystyle k,n\in \mathbb N$.Thank you very much!

All the best,

Integrator

October 29th, 2019, 10:07 AM   #6
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 Originally Posted by Integrator Some say that $\displaystyle \sum_{k=n}^{k=1} k$ it's not equal with $\displaystyle \sum_{k=1}^{k=n} k$ where $\displaystyle k,n\in \mathbb N$.
Well, who says that, and what do they say about it?

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
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 Originally Posted by Integrator "Some say" that $\displaystyle \sum_{k=n}^{k=1} k$ it's not equal with $\displaystyle \sum_{k=1}^{k=n} k$ where $\displaystyle k,n\in \mathbb N$.
Sounds like one of Trump's opening lines ...

October 29th, 2019, 11:22 AM   #8
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 Originally Posted by skeeter Sounds like one of Trump's opening lines ...
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
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 Originally Posted by DarnItJimImAnEngineer Well, who says that, and what do they say about it? 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.
Hello,

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
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Quote:
 Originally Posted by Integrator which means that
Why?

I can't see how you could justify the term $\displaystyle \frac{n + 1}{2}$ in your first equation in such a way that it would still be justified in the second equation.

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