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Anonymous Sun 22 Dec 16:19:16 2019 No.11245765 View Reply Original Report

Quoted By: >>11245770 >>11245784 >>11246328 >>11246739 >>11247454 >>11249535 >>11249780

is there some kind of formula to sum numbers in a row
like 1+2+3+4+5+6+7+8+9+10+11 etc.
I'm struggling to come up with one and if there's already one I'd like not to spend my evening trying to find one.

Anonymous

Anonymous Sun 22 Dec 2019 16:22:05 No.11245768 Report

Quoted By:

Take the average and multiply by the quantity of numbers. The average is the average of the smallest number and biggest number. (1+11)/2 * 11 = 66

Anonymous

Anonymous Sun 22 Dec 2019 16:24:33 No.11245770 Report

Quoted By:

>>11245765
S = 1+2 +...+n
S = n+(n-1)+...+1
2S = n(n+1)
S = n(n+1)/2

Anonymous

Anonymous Sun 22 Dec 2019 16:37:37 No.11245784 Report

Quoted By:

>>11245765
Take your finite series
Represent it as a bar graph, where the horizontal axis represents index, and the vertical axis represents value
Rotate the graph 180 degrees
Place the rotated graph above your original graph, so that the columns correspond to each other, in their respective reverse orders
The figure you have is a rectangle, with width equal to the cardinality of your indices, and height equal to the largest value plus the smallest value.

Anonymous

Anonymous Sun 22 Dec 2019 21:23:43 No.11246328 Report

Quoted By: >>11246351

>>11245765
summation of an arithmetic series
should have been covered in pre-calculus

Anonymous

Anonymous Sun 22 Dec 2019 21:40:09 No.11246351 Report

Quoted By:

>>11246328
total/2(total+increment)

Anonymous

Anonymous Mon 23 Dec 2019 01:21:21 No.11246739 Report

Quoted By:

>>11245765
Yes its called the gauss trick.
The sum of any set of numbers in a row is equal to S =n(n+1) divided by 2.

Anonymous

Anonymous Mon 23 Dec 2019 07:54:15 No.11247454 Report

Quoted By: >>11248705

>>11245765
Retard or underage?

Anonymous

Anonymous Mon 23 Dec 2019 20:02:44 No.11248705 Report

Quoted By:

>>11247454
yes

Anonymous

Anonymous Tue 24 Dec 2019 03:28:06 No.11249535 Report

Quoted By: >>11249749

>>11245765
>tfw gauss found out the formula on his own when he was 15 in 1500
>he was literally smarter than his class and teacher combined

why live

Anonymous

Anonymous Tue 24 Dec 2019 05:36:30 No.11249749 Report

Quoted By:

>>11249535
>in 1500
This was in the 1790s

I'm sure if you made it your goal to find that formula having never seen it before you could do it too. Just not as quickly

Anonymous

Anonymous Tue 24 Dec 2019 05:53:39 No.11249780 Report

Quoted By:

>>11245765
$\displaystyle \sum_{n=1}^N n = \frac{N}{2}(N+1)$

Ill give you a better one too

$\displaystyle \sum_{n=0}^{N} (2n+1) = N^2$

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