Comments on: Math problem

Question: Math problem

Kwantsar — Sat, 08 Oct 2005 11:04:42 -0800

A crystal consists of 100,000,000 layers of atoms such that there is 1 atom in the first layer, 3 in the second, 6 in the third, 10 in the fourth, 15 in the fifth, and so forth. Exactly how many atoms are there in the entire crystal?

Computer-aided solutions are not helpful (post them anyway, If you want to show off, I don't really give a damn). There's gotta be a way to solve this by hand (without using super-fancy-math-shit), right? What is it?

By: vacapinta

vacapinta — Sat, 08 Oct 2005 11:20:29 -0800

This series is known as the triangular numbers. The sum of any N triangular numbers produces a new series known as the tetrahedral numbers. Anyways, you want the nth tetrahedral number where N=10^8

S=(n)(n+1)(n+2)/6

which is about 10^23 if im not mistaken.

By: Kwantsar

Kwantsar — Sat, 08 Oct 2005 11:27:02 -0800

Awesomeness.

By: RichardP

RichardP — Sat, 08 Oct 2005 11:32:53 -0800

Assuming you intend the sequence defined by:

a₀ = 0
a_n = a_n-1 + n

We can solve for the generator, and get:

a_n = n(n+1)/2

Next, we want to find the generator for:

b₀ = a₀
b_n = b_n-1 + a_n

Solving, we get:

b_n = (n+1)(n+2)(n+3)/6

Plugging in 100,000,000 for n, we get:

100000001*100000002*100000003/6

which is...

166666676666666850000001 atoms

No computer aided solutions or fancy math shit needed.

By: Kwantsar

Kwantsar — Sat, 08 Oct 2005 11:34:21 -0800

More awesomeness.

By: RichardP

RichardP — Sat, 08 Oct 2005 11:37:31 -0800

Oops, that last bit should have been as follows:

Solving, we get:

b_n = n(n+1)(n+2)/6

Plugging in 100,000,000 for n, we get:

100000000*100000001*100000002/6

which is...

166666671666666700000000 atoms

By: shmegegge

shmegegge — Sat, 08 Oct 2005 12:29:58 -0800

i know this isn't a solution, but i love this askme question, and kwantsar's replies to the excellent answers he's gotten. sorry for the derail.

By: notsnot

notsnot — Sat, 08 Oct 2005 15:21:15 -0800

how'd you get from the definition of bn = bn-1 + an thru to "solving..."?

By: gleuschk

gleuschk — Sun, 09 Oct 2005 14:02:44 -0800

This may not be what RichardP did, but here's one way to go: rewrite it as
b_n = Σ_0..n k(k+1)/2.

Then break up the interior and use standard summation formulas: the sum of the first n integers is n(n+1)/2 (already used above) and the sum of the first n squares is n(n+1)(2n+1)/6.