Math problem

October 8, 2005 11:04 AM

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?

posted by Kwantsar to grab bag (8 comments total)

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.
posted by vacapinta at 11:20 AM on October 8, 2005

Awesomeness.
posted by Kwantsar at 11:27 AM on October 8, 2005

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.
posted by RichardP at 11:32 AM on October 8, 2005

More awesomeness.
posted by Kwantsar at 11:34 AM on October 8, 2005

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

posted by RichardP at 11:37 AM on October 8, 2005

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.
posted by shmegegge at 12:29 PM on October 8, 2005

how'd you get from the definition of bn = bn-1 + an thru to "solving..."?
posted by notsnot at 3:21 PM on October 8, 2005

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.
posted by gleuschk at 2:02 PM on October 9, 2005