document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1118782>Question 1118782</A>: <I><SPAN STYLE=\"word-break: break-all;\"> sum from n = 1 to n = infinity(2^n - 1)/3^n </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #734219 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=734219\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "S = (2^1-1)/3^1 + (2^2-1)/3^2 + {2^3-1/3^3 + ...<br><BR>\n" );
document.write( "S = 1/3 + 3/9 + 7/27 + 15/81 + ...<br><BR>\n" );
document.write( "If the denominators of the series form a geometric progression but the whole series is not geometric, then multiply that expression for the sum by the common ratio; then subtract the two series.<br><PRE STYLE=\"white-space: pre-wrap; white-space: -moz-pre-wrap; white-space: -pre-wrap; white-space: -o-pre-wrap; word-wrap: break-word;\">\r\n" );
document.write( "  3S = 1 + 3/3 + 7/9 + 15/27 + ...\r\n" );
document.write( "   S =     1/3 + 3/9 + 7/27 + ...\r\n" );
document.write( "  --------------------------------\r\n" );
document.write( "  2S = 1 + 2/3 + 4/9 + 8/27 + ...</PRE><br><BR>\n" );
document.write( "The expression on the right is a pure geometric series...<br><BR>\n" );
document.write( "  2S = 1/(1-(2/3)) = 1/(1/3) = 3<BR>\n" );
document.write( "   S = 3/2 or 1.5 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );