document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185725>Question 1185725</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this <BR>\n" );
document.write( "1-((2-1)/4)-((2^2-1)/4^2)-((2^3-1)/4^3)-((2^4-1)/4^4)- infinite </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #816601 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=816601\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "1 - ((2-1)/4) - ((2^2-1)/4^2) - ((2^3-1)/4^3) - ((2^4-1)/4^4) - ...<br><BR>\n" );
document.write( "1 - 1/4 - (4-1)/16 - (8-1)/64 - (16-1)/256 - ...<br><BR>\n" );
document.write( "3/4 - (4/16-1/16) - (8/64-1/64) - (16/256-1/256) - ...<br><BR>\n" );
document.write( "3/4 - (4/16+8/64+16/256...) + (1/16+1/64+1/256...)<br><BR>\n" );
document.write( "The two expressions in parentheses are infinite geometric sequences with common ratio less than 1.<br><BR>\n" );
document.write( "4/16+8/64+16/256+... = 1/4+1/8+1/16+... = (1/4)/(1-1/2) = (1/4)/(1/2) = (1/4)(2/1) = 1/2<br><BR>\n" );
document.write( "1/16+1/64+1/256+... = (1/16)/(1-1/4) = (1/16)/(3/4) = (1/16)(4/3) = 1/12<br><BR>\n" );
document.write( "The sum of the sequence is then<br><BR>\n" );
document.write( "3/4 - 1/2 + 1/12 = 9/12 - 6/12 + 1/12 = 4/12 = 1/3<br><BR>\n" );
document.write( "ANSWER: 1/3<br><BR>\n" );
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