/sci/ - Science & Math

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

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

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

[math]
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)
[/math]

[math]
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)
[/math]

>>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

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

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

>>11976694
>>11976667
He was 3.

>>11976667
[math]\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 [/math]

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

>>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).

>>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.

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