/sci/ - Science & Math

[Anonymous No.2451836]

That is to say, 1(n)+2(n-1)+3(n-2)+...+(n-2)3+(n-1)2+(1)n
Can this be simplified, like 1+2+...+n=n(n+1)/2 ? If you find an answer, I'd love to know how you got it. I'll post some things I've noticed about this in case it helps.

[Anonymous No.2451857]

This statement is equivalent to the statement
(1)+(1+2)+(1+2+3)+(1+2+3+4)+...+(1+2+...+n)
Parenthesis have been added to make the pattern obvious.

[Anonymous No.2451940]

x:f(x)
------
1 : 1
2 : 4
3 : 10
4 : 20
5 : 35
6 : 56
7 : 84

[Anonymous No.2451989]

>>2451857
<div class="math">\displaystyle \sum_{k=1}^n \left(\sum_{i=1}^k i\right) = \sum_{k=1}^n\frac{k(k+1)}{2} = \frac{1}{2}\sum_{k=1}^n k^2 + \frac{1}{2}\sum_{k=1}^n k = \frac{1}{2}\frac{n(n+1)(2n+1)}{6} + \frac{1}{2}\frac{n(n+1)}{2}</div>

[Anonymous No.2452029]

>>2451989
Thank you anon! If it interests you, I'm using this information coupled with Fibonacci cubes in an attempt to redefine Fibonacci numbers.

[Anonymous No.2452128]

>>2451989
\displaystyle = \frac{n(n+1)}{4}(\frac{2n+1}{3} + 1) = \frac{n(n+1)}{4}(\frac{2n+1}{3} + \frac{3}{3}) = \frac{n(n+1)}{4}(\frac{2n+4}{3}) = \frac{n(n+1)}{2}(\frac{n+2}{3}) = \frac{n(n+1)(n+2)}{6}

[Anonymous No.2452148]

>>2452128
Well I don't know exactly what I'm doing, but my final answer there was n(n+1)(n+2)/6, which holds for all of >>2451940
Today is a good day.

[Anonymous No.2452229]

WTF is going on in this thread?

[Anonymous No.2452265]

>>2452229
A lot of samefagging by me, but also quite a bit of summation.

[Anonymous No.2452301]

>>2452148
https://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity#Hockey-Stick_Identity

[Anonymous No.2452357]

>>2451989
Sorry to intrude on your math and also be a newfag, but how did you post with LaTeX on 4chan?

[Anonymous No.2452408]

>>2452357
Use the little jsMath button at the bottom right hand corner?

[Anonymous No.2452450]

Since you seem well versed in sums, can you tell me how you got n(n+1)(2n + 1) / 6? To be more specific, I see a similar formula with riemann sums but I never quite got the intuition behind it. I'm not the smarted guy, so I'm probably missing something obvious.

[Anonymous No.2452484]

>>2452450
I actually had found this one before I had a chance to try and derive it myself, so I can't tell you that. I can, however, explain why the division by 6 always yields an integer and I can prove it by induction, but that's the extent of my knowledge there.

>>2452301
I THOUGHT I saw something close to Pascal's triangle in the numbers, but when you're trying to write everything down on the glass of a steamy shower, reference materials are pretty difficult to keep. But I digress.

[F(n)=xth Fibonacci number No.2452533]

F(n+2)=f(n)

I've been working withe these numbers, and I was wondering if an obvious pattern immediately jumped out at anyone:

n : F(n+2) : values for each possible amount of 1's
3 : 5 : 1+n+(n-2){1}
4 : 8 : 1+n+(n-2)(n-1)/2{1}
5 : 13 : 1+n+(n-2)(n-1)/2+(n-4){1}
6 : 21 : 1+n+(n-2)(n-1)/2+(n-4)(n-3)/2+{n-5}{1}
7 : 34 : 1+n+(n-2)(n-1)/2+(n-4)(n-3)(n-2)/6+(n-6)(n-5)(n-4)/6{1}
8 : 55 : 1+n+(n-2)(n-1)/2+(n-3)(n-2)(n-1)/6+(n-5)(n-4)(n-3)/6+(n-7)(n-6)(n-5)/6{1}

If anyone here is unfamiliar with Fibonacci Cubes, you take a string of n digits of 1's and 0's and create all possible strings where no two 1's are touching. I've arranged my terms so that each term represents each amount of 1's used. The {1} at the end of each line is to show that each final term equals 1.

[Anonymous No.2452654]

Bump.

[Anonymous No.2452902]

Bump.