document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=572482>Question 572482</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1(1!) + 2(2!) + 3(3!) + . . . + n(n!).\r<BR>\n" );
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document.write( "Find the sum of this series if n = 15. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #368409 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=368409\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1(1!) + 2(2!) + ... + 15(15!)\r<BR>\n" );
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document.write( "15(15!) can be rewritten as (16-1)15!, or 16! - 15!. Similarly, 14(14!) can be rewritten as 15! - 14!. This pattern continues and we can rewrite 1(1!) as 2! - 1!. Adding these up, we obtain\r<BR>\n" );
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document.write( "(2! - 1!) + (3! - 2!) + ... + (16! - 15!)\r<BR>\n" );
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document.write( "All the terms cancel, except the -1! and 16!. Therefore the series telescopes to 16! - 1!, or 16! - 1. </SPAN>\n" );
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