document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1062323>Question 1062323</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, please help me turn this series into summation notation. Thank you in advance.\r<BR>\n" );
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document.write( "1. 2!/7 + 3!/14 + 4!/21 + ... + (n-1)!/7n\r<BR>\n" );
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document.write( "2. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+-1%2F5+%2B+8%2F25+-+27%2F125+%2B+64%2F625+ ALIGN=MIDDLE  ALT=\"+-1%2F5+%2B+8%2F25+-+27%2F125+%2B+64%2F625+\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #677185 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=677185\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I like the last term's being (n+1)!<BR>\n" );
document.write( "Then I can write sum from n=1 to n of (n+1)!/7n.  Is the last term (n+1)!?<BR>\n" );
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document.write( "The second term is the sum from n=1 to 4 of -1^(n)*(n^3)/5^(n).<BR>\n" );
document.write( "The -1^(n) will alternately be negative-positive, beginning with negative. </SPAN>\n" );
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