document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1036461>Question 1036461</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Compute the sum\r<BR>\n" );
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document.write( "1 + 1/3 + 2/3 + 2/9 + 4/9 + 4/27 + 8/27 + 8/81...\r<BR>\n" );
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document.write( "Thanks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #651134 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=651134\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it looks like you have 2 geometric sequences going here.\r<BR>\n" );
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document.write( "the first geometric sequence is composed of all odd number terms.\r<BR>\n" );
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document.write( "1 + 2/3 + 4/9 + 8/27 ...\r<BR>\n" );
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document.write( "the common ratio would be 2/3.\r<BR>\n" );
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document.write( "1 * 2/3 = 2/3 * 2/3 = 4/9 * 2/3 = 8/27, ...\r<BR>\n" );
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document.write( "the second geometric sequence is composed of all even number terms.\r<BR>\n" );
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document.write( "1/3 + 2/9 + 4/27 + ...\r<BR>\n" );
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document.write( "the common ratio is again 2/3.\r<BR>\n" );
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document.write( "1/3 * 2/3 = 2/9 * 2/3 = 4/27, ...\r<BR>\n" );
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document.write( "if you take the sum of both geometric series and add them together, you should get the sum of the combined series.\r<BR>\n" );
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document.write( "you can test this out by taking the sum of each series to the 3d term.\r<BR>\n" );
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document.write( "this will show you that the formula works.\r<BR>\n" );
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document.write( "once you are assured of that, you can then take the sum to infinity.\r<BR>\n" );
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document.write( "the sum of a geometric series is Sn = A1 * (1 - r^n) divided by (1 - r).\r<BR>\n" );
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document.write( "in the first geometric series, we would get S3 = 1 * (1 - (2/3)^3) divided by (1 - 2/3) which would be equal to 1 * 2.11111\r<BR>\n" );
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document.write( "add 1 and 2/3 and 4/9 together to get 2.111...\r<BR>\n" );
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document.write( "the formula works for the first geometric series.\r<BR>\n" );
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document.write( "in the second geometric series, we would get S3 = 1/3 * (1 - 2/3)^3) divided by (1 - 2/3) which would be equal to .7037037037.\r<BR>\n" );
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document.write( "add 1/3 and 2/9 and 4/27 together to get .7037037037.\r<BR>\n" );
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document.write( "the formula works for the second geometric series.\r<BR>\n" );
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document.write( "the first 3 terms of the first geometric series added to the first 3 terms of the second geometric series gets you.\r<BR>\n" );
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document.write( "1 + 1/3 + 2/3 + 2/9 + 4/9 + 4/27\r<BR>\n" );
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document.write( "add these together and you get 2.814814815\r<BR>\n" );
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document.write( "add 2.1111111111 and .7037037037. together and you get 2.814814815.\r<BR>\n" );
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document.write( "the formula works for the combined series.\r<BR>\n" );
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document.write( "now that you know it works, you can solve for the infinite series.\r<BR>\n" );
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document.write( "the formula for the infinite series is S&infin; = A1 * 1 / (1-r)\r<BR>\n" );
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document.write( "since 1/(1-r) = 1/(1-2/3) = 1/(1/3) = 3, you would get 1 * 3 + 1/3 * 3 = 4.\r<BR>\n" );
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document.write( "the sum of the infinite series is 4.\r<BR>\n" );
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