# Math problem

October 8, 2005 11:04 AM Subscribe

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?

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?

Best answer: Assuming you intend the sequence defined by:

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

aWe can solve for the generator, and get:_{0}= 0

a_{n}= a_{n-1}+ n

aNext, we want to find the generator for:_{n}= n(n+1)/2

bSolving, we get:_{0}= a_{0}

b_{n}= b_{n-1}+ a_{n}

bPlugging in 100,000,000 for n, we get:_{n}= (n+1)(n+2)(n+3)/6

100000001*100000002*100000003/6which is...

166666676666666850000001 atomsNo computer aided solutions or fancy math shit needed.

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

Oops, that last bit should have been as follows:

Solving, we get:

Solving, we get:

bPlugging in 100,000,000 for n, we get:_{n}= n(n+1)(n+2)/6

100000000*100000001*100000002/6which is...

166666671666666700000000 atomsposted 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

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

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

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

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

This thread is closed to new comments.

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