/sci/ - Science & Math » Thread #11976667

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Anonymous Wed 05 Aug 12:52:39 2020 No.11976667 View Reply Original Report

Quoted By: >>11976677 >>11976710 >>11977584 >>11977785 >>11978066

Don't cheat, show your method :) A 5-year-old Gauss solved this in kindergarten.

? from 1 to 100 x

Anonymous

Anonymous Wed 05 Aug 2020 12:56:39 No.11976670 Report

Quoted By: >>11976694

5050.

Put 1 and 99 together to make 100. Repeat this 49 times. Add the 50 from the middle, and the final 100.
49 x 100 + 50 + 100 = 5050

Anonymous

Anonymous Wed 05 Aug 2020 13:00:46 No.11976677 Report

Quoted By:

>>11976667
(x(x+1)) / 2 :)

Anonymous

Anonymous Wed 05 Aug 2020 13:09:36 No.11976694 Report

Quoted By: >>11977584

>>11976670
crazy that a 5 yo could solve that :0

Anonymous

Anonymous Wed 05 Aug 2020 13:11:22 No.11976697 Report

Quoted By:

$ S=1+2+ \cdots +n \\ S=n+(n-1)+ \cdots +1 \\ 2S=1+n+2+(n-1)+ \cdots n+1=n(n+1) \\ S=\frac{n}{2}(n+1) $

Anonymous

Anonymous Wed 05 Aug 2020 13:12:37 No.11976700 Report

Quoted By:

$ S=1+2+ \cdots +n \\ S=n+(n-1)+ \cdots +1 \\ 2S=1+n+2+(n-1)+ \cdots +n+1=n(n+1) \\ S=\frac{n}{2}(n+1) $

Anonymous

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Anonymous Wed 05 Aug 2020 13:16:43 No.11976710 Report

Quoted By: >>11976718

>>11976667
sum 1 to a where a is even : (a-0)+1 + (a-1)+2 + ... + ((a/2)+(a/2+1)) = (a/2)*(a/2+1)
sum 1 to 100 : 100+1 + 101+2... + 50+51 = 101*50 = 5050

Anonymous

Anonymous Wed 05 Aug 2020 13:18:41 No.11976718 Report

Quoted By: >>11976725

>>11976710
not (a/2)*(a/2+1) but (a+1)*(a/2) fuck

Anonymous

Anonymous Wed 05 Aug 2020 13:20:04 No.11976725 Report

Quoted By:

>>11976718
great, now prove that every positive integer is the sum of at most 3 triangular numbers (proven by gauss :) don't cheat :)

Anonymous

Anonymous Wed 05 Aug 2020 18:07:35 No.11977584 Report

Quoted By:

>>11976694
>>11976667
He was 3.

Anonymous

Anonymous Wed 05 Aug 2020 18:51:29 No.11977785 Report

Quoted By: >>11977799

>>11976667
$\int_{0}^{1}f(x)dx = [f(x)(x+c)]_{0}^{1} - \int_{0}^{1} f'(x)(x+c)dx.\\ \int_{1}^{100}xdx = \sum\limits_{n=1}^{99}\int_{0}^{1}(x+n)dx = \sum\limits_{n=1}^{99}([(x+n)(x+c)]_{0}^{1} - \int_{0}^{1} 1(x+c)dx)\\ let\ c=-1/2.\\ \int_{1}^{100}xdx = 1/2 +100/2+\sum\limits_{n=2}^{99}n\\ (100^2/2 - 1/2)+1/2+100/2=\sum\limits_{n=1}^{100}n$

Anonymous

Anonymous Wed 05 Aug 2020 18:55:54 No.11977799 Report

Quoted By: >>11977817

>>11977785
i'bite, what is c is not -1/2

Anonymous

Anonymous Wed 05 Aug 2020 18:59:54 No.11977817 Report

Quoted By:

>>11977799
the integral on the right doesn't vanish so you end up with a -99*const term and the 1/2 weights on the 1 and 100 are different (they still sum to 1).

Anonymous

Anonymous Wed 05 Aug 2020 20:07:28 No.11978066 Report

Quoted By:

>>11976667
Did Gauss really solve this PDE at age 5? I find it hard to believe, even for one of the greatest geniuses of math.

Anonymous

Anonymous ## Mod Thu 06 Aug 2020 12:46:04 No.11980415 Report

Quoted By:

This thread was moved to >>>/wsr/867964

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