document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=259612>Question 259612</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write the geometric series -9/2 + 3/2 - 1/2 + 1/6- ... + 1/39366 in summation notation\r<BR>\n" );
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document.write( "Anyone?<BR>\n" );
document.write( "(: </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #191073 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=191073\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> google \"sum of a geometric series\"\r<BR>\n" );
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document.write( "the ratio (between consecutive terms) is -1/3\r<BR>\n" );
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document.write( "you can find the number of terms in the series from ___ -1/2 r^k = \"last term\"<BR>\n" );
document.write( "___ where k is the number of terms beyond -1/2 and r is -1/3<BR>\n" );
document.write( "___ k should be around 10 </SPAN>\n" );
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