document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1160668>Question 1160668</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the sum of the infinite geometric series \r<BR>\n" );
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document.write( "3/5-1/5+1/15-1/45+...\r<BR>\n" );
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document.write( "1+1/4+1/16+...\r<BR>\n" );
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document.write( "64+16+4+...\r<BR>\n" );
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document.write( "1/4+1/16+1/64+...\r<BR>\n" );
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document.write( "The following problem refers to a geometric sequence:<BR>\n" );
document.write( "If a_1= -1 and r= -1, find a_45\r<BR>\n" );
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document.write( "If a_1= 3 and r= 5, find a_n\r<BR>\n" );
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document.write( "Determine if the following sequence is geometric progressions. If it is, identify the common ratio r, if <BR>\n" );
document.write( "not answer DNE.\r<BR>\n" );
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document.write( "1/5,-1/15, 1/45, -1/135,...\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #784031 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=784031\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll do the first two problems to get you started.\r<BR>\n" );
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document.write( "Problem 1\r<BR>\n" );
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document.write( "The first term is a = 3/5<BR>\n" );
document.write( "The common ratio is r = -1/3\r<BR>\n" );
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document.write( "We multiply each term by -1/3 to get the next term<BR>\n" );
document.write( "first term = -3/5<BR>\n" );
document.write( "second term = (common ratio)*(first term) = (-1/3)*(3/5) = -1/5<BR>\n" );
document.write( "third term = (common ratio)*(second term) = (-1/3)*(-1/5) = 1/15<BR>\n" );
document.write( "etc etc\r<BR>\n" );
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document.write( "With r = -1/3 = -0.33 (approximately), we can see that -1 < r < 1 is true, which means that the infinite geometric sum converges to a finite number. That number is...<BR>\n" );
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document.write( "Answer: <font color=red size=4> 9/20 </font>\r<BR>\n" );
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document.write( "With your calculator, note how <BR>\n" );
document.write( "<ul><li>3/5-1/5 = 0.4</li><li>3/5-1/5+1/15 = 0.46666666666667 (approximate)</li><li>3/5-1/5+1/15-1/45 = 0.44444444444444 (approximate)</li><li>3/5-1/5+1/15-1/45+1/135 = 0.45185185185186 (approximate)</li><li>3/5-1/5+1/15-1/45+1/135-1/405 = 0.44938271604939 (approximate)</li><li>3/5-1/5+1/15-1/45+1/135-1/405+1/1215 = 0.45020576131688 (approximate)</li></ul>Those partial sums are slowly approaching 9/20 = 0.45; they'll never actually get there because we can't actually reach infinity. The sums only get closer and closer.\r<BR>\n" );
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document.write( "Problem 2\r<BR>\n" );
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document.write( "Same idea as problem 1. We have different values this time of course. In this problem, <BR>\n" );
document.write( "a = 1 is the first term<BR>\n" );
document.write( "r = 1/4 is the common ratio\r<BR>\n" );
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document.write( "We can determine the common ratio by picking any term but the first one, then dividing it over its previous term\r<BR>\n" );
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document.write( "examples:<BR>\n" );
document.write( "(second term)/(first term) = (1/4) divided by (1) = 1/4<BR>\n" );
document.write( "(third term)/(second term) = (1/16) divided by (1/4) = (1/16)*(4/1) = 4/16 = 1/4<BR>\n" );
document.write( "These are two of many ways to see why the common ratio is r = 1/4.\r<BR>\n" );
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document.write( "The infinite sum converges to a finite value because r = 1/4 = 0.25 makes -1 < r < 1 true. \r<BR>\n" );
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document.write( "Side note: We can condense <font color=blue>-1 < r < 1</font> into the absolute value inequality <font color=blue> |r| < 1 </font>. This says r is within 1 unit of zero, or that the distance from r to 0 is less than 1.\r<BR>\n" );
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document.write( "Anyways, back to the problem. We plug a = 1 and r = 1/4 into the formula used in the prior question.\r<BR>\n" );
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document.write( "Answer: <font color=red size=4> 4/3 </font>\r<BR>\n" );
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document.write( "Partial check or partial confirmation we have the right answer:<BR>\n" );
document.write( "<ul><li>1+1/4 = 1.25</li><li>1+1/4+1/16 = 1.3125</li><li>1+1/4+1/16+1/64 = 1.328125</li><li>1+1/4+1/16+1/64+1/256 = 1.33203125</li><li>1+1/4+1/16+1/64+1/256+1/1024 = 1.3330078125</li><li>1+1/4+1/16+1/64+1/256+1/1024+1/4096 = 1.333251953125</li></ul>The partial sums are gradually approaching 4/3 = 1.33333...\r<BR>\n" );
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