document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1190304>Question 1190304</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this math problem:If the sum of 1+2x+4x²+8x³+... 1¼. Find x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #821919 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=821919\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "first term = 1<BR>\n" );
document.write( "second term = 2x = 2x*(First term)<BR>\n" );
document.write( "third term = 4x^2 = 2x*2x = 2x*(second term)<BR>\n" );
document.write( "fourth term = 8x^3 = 2x*4x^2 = 2x*(third term)<BR>\n" );
document.write( "nth term = 2x*(term just before nth term)\r<BR>\n" );
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document.write( "As you can see, we have a geometric sequence with: <BR>\n" );
document.write( "a = 1 = first term<BR>\n" );
document.write( "r = 2x = common ratio<BR>\n" );
document.write( "We start at 1, and each time we need a new term, we multiply by 2x.\r<BR>\n" );
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document.write( "If we want the infinite geometric series of 1+2x+4x^2+8x^3+... to converge to some finite value, then we need -1 < r < 1 to be true.<BR>\n" );
document.write( "r must be between -1 and 1, excluding both endpoints.<BR>\n" );
document.write( "Why? because we need to add smaller and smaller pieces in order to slowly approach the finite sum. <BR>\n" );
document.write( "Otherwise, the sum will diverge and blow up to plus or minus infinity.<BR>\n" );
document.write( "Or it may bounce around never settling on any value at all. \r<BR>\n" );
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document.write( "-1 < r < 1<BR>\n" );
document.write( "-1 < 2x < 1<BR>\n" );
document.write( "-1/2 < x < 1/2<BR>\n" );
document.write( "-0.5 < x < 0.5<BR>\n" );
document.write( "This establishes the boundaries of what x can be. \r<BR>\n" );
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document.write( "If -1 < r < 1 is true, then we can use this summation formula<BR>\n" );
document.write( "S = a/(1-r)<BR>\n" );
document.write( "where S in this case is the result of adding up the infinitely many terms of the geometric series.\r<BR>\n" );
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document.write( "Plug in the first term a = 1, the common ratio r = 2x, and the desired sum S = 1 & 1/4 = 1 + 1/4 = 1.25 and solve for x.\r<BR>\n" );
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document.write( "S = a/(1-r)<BR>\n" );
document.write( "1.25 = 1/(1-2x)<BR>\n" );
document.write( "1.25(1-2x) = 1<BR>\n" );
document.write( "1.25-2.5x = 1<BR>\n" );
document.write( "1.25-1 = 2.5x<BR>\n" );
document.write( "2.5x = 0.25<BR>\n" );
document.write( "x = 0.25/2.5<BR>\n" );
document.write( "x = 25/250<BR>\n" );
document.write( "x = (25*1)/(25*10)<BR>\n" );
document.write( "x = 1/10<BR>\n" );
document.write( "x = 0.1<BR>\n" );
document.write( "Now let's go back to -0.5 < x < 0.5<BR>\n" );
document.write( "Notice that x = 0.1 is in this interval, so we have satisfied the criteria needed for an infinite geometric sum.\r<BR>\n" );
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document.write( "Answer: <font color=red>x = 1/10</font> or <font color=red>x = 0.1</font><BR>\n" );
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