document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1012259>Question 1012259</A>: <I><SPAN STYLE=\"word-break: break-all;\"> So.. for my holiday homework i was told to find a formula to sum up the tetrahedral numbers.. which are 1, 4, 10, 20, 35, 56...\r<BR>\n" );
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document.write( "And im really stuck on where to start, could you please help me step by step to sum up consecutive tetrahedral numbers? (Would appreciate if its in a easy mathematical language)\r<BR>\n" );
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document.write( "Is there any tricks to find formulas or equstions from a sequence?\r<BR>\n" );
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document.write( "Thank you :\"c </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #628139 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=628139\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The tetrahedral numbers are found in Pascal's triangle, namely<BR>\n" );
document.write( "Tn = combination of (n+2) taken 3 at a time = (n+2)! / (3! * (n+2-3)!) =<BR>\n" );
document.write( "(n+2)! / (3! * (n-1)!) where Tn is the nth tetrahedral number<BR>\n" );
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document.write( "another way to solve this is to note that the sum of the first n triangular numbers is the nth tetrahedral number,that is<BR>\n" );
document.write( "Tn = (n(n+1)(n+2)) / 6<BR>\n" );
document.write( "a triangular number can be thought of as the number of dots needed to form an equilateral triangle, namely<BR>\n" );
document.write( "1, 3, 6, 10, 15, 21, and so forth<BR>\n" );
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document.write( "the nth sum(Sn) of n consecutive tetrahedral numbers is<BR>\n" );
document.write( "Sn = (n(n+2)(n+3)) / 24<BR>\n" );
document.write( "We can derive this formula by starting with<BR>\n" );
document.write( "Tn = (n(n+1)(n+2)) / 6, then expand the numerator<BR>\n" );
document.write( "Tn = (n^3 + 3n^2 + 2n) / 6<BR>\n" );
document.write( "Next we use the formulas for the sum of first n consecutive integers, first n squared integers and first n cubed integers, namely<BR>\n" );
document.write( "Sn = (n(n+1)) / 2<BR>\n" );
document.write( "Sn^2 = (n(n+1)(2n+1)) / 6<BR>\n" );
document.write( "Sn^3 = (n^2(n+1)^2 / 4<BR>\n" );
document.write( "Substitute these equations into our Tn formula and simplify <BR>\n" );
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