Comments on: Math problem
http://ask.metafilter.com/25201/Math-problem/
Comments on Ask MetaFilter post Math problemSat, 08 Oct 2005 11:20:29 -0800Sat, 08 Oct 2005 11:20:29 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Math problem
http://ask.metafilter.com/25201/Math-problem
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? <br /><br /> 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?post:ask.metafilter.com,2005:site.25201Sat, 08 Oct 2005 11:04:42 -0800KwantsarpuzzlequizmindbenderexpansionmathmathmaticsmathsBy: vacapinta
http://ask.metafilter.com/25201/Math-problem#398596
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<br>
<br>
S=(n)(n+1)(n+2)/6<br>
<br>
which is about 10^23 if im not mistaken.comment:ask.metafilter.com,2005:site.25201-398596Sat, 08 Oct 2005 11:20:29 -0800vacapintaBy: Kwantsar
http://ask.metafilter.com/25201/Math-problem#398600
Awesomeness.comment:ask.metafilter.com,2005:site.25201-398600Sat, 08 Oct 2005 11:27:02 -0800KwantsarBy: RichardP
http://ask.metafilter.com/25201/Math-problem#398604
Assuming you intend the sequence defined by:<blockquote>a<sub>0</sub> = 0<br>
a<sub>n</sub> = a<sub>n-1</sub> + n</blockquote>We can solve for the generator, and get:<blockquote>a<sub>n</sub> = n(n+1)/2</blockquote>Next, we want to find the generator for:<blockquote>b<sub>0</sub> = a<sub>0</sub><br>
b<sub>n</sub> = b<sub>n-1</sub> + a<sub>n</sub></blockquote>Solving, we get:<blockquote>b<sub>n</sub> = (n+1)(n+2)(n+3)/6</blockquote>Plugging in 100,000,000 for n, we get:<blockquote>100000001*100000002*100000003/6</blockquote>which is...<blockquote>166666676666666850000001 atoms</blockquote>No computer aided solutions or fancy math shit needed.comment:ask.metafilter.com,2005:site.25201-398604Sat, 08 Oct 2005 11:32:53 -0800RichardPBy: Kwantsar
http://ask.metafilter.com/25201/Math-problem#398606
More awesomeness.comment:ask.metafilter.com,2005:site.25201-398606Sat, 08 Oct 2005 11:34:21 -0800KwantsarBy: RichardP
http://ask.metafilter.com/25201/Math-problem#398608
Oops, that last bit should have been as follows:<br>
<br>
Solving, we get:<blockquote>b<sub>n</sub> = n(n+1)(n+2)/6</blockquote>Plugging in 100,000,000 for n, we get:<blockquote>100000000*100000001*100000002/6</blockquote>which is...<br>
<blockquote>166666671666666700000000 atoms </blockquote>comment:ask.metafilter.com,2005:site.25201-398608Sat, 08 Oct 2005 11:37:31 -0800RichardPBy: shmegegge
http://ask.metafilter.com/25201/Math-problem#398634
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.comment:ask.metafilter.com,2005:site.25201-398634Sat, 08 Oct 2005 12:29:58 -0800shmegeggeBy: notsnot
http://ask.metafilter.com/25201/Math-problem#398740
how'd you get from the definition of bn = bn-1 + an thru to "solving..."?comment:ask.metafilter.com,2005:site.25201-398740Sat, 08 Oct 2005 15:21:15 -0800notsnotBy: gleuschk
http://ask.metafilter.com/25201/Math-problem#399182
This may not be what RichardP did, but here's one way to go: rewrite it as <br>
b<sub>n</sub> = Σ</big></big><sub>0..n</sub> k(k+1)/2.<br>
<br>
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.comment:ask.metafilter.com,2005:site.25201-399182Sun, 09 Oct 2005 14:02:44 -0800gleuschk