document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=619397>Question 619397</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I can do this when the numbers are whole, but fractions are really throwing me off. Thank you for any help you can provide!!\r<BR>\n" );
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document.write( "Write the geometric series -9/2+3/2-1/2+1/6-...+1/39366 in summation notation. Then using the formula for the sum of a geometric series, compute the sum. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #389633 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=389633\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the formulas for a geometric progression are:<BR>\n" );
document.write( "L = a*r^(n-1)<BR>\n" );
document.write( "L = the last term in the sequence.<BR>\n" );
document.write( "a = the first term in the sequence.<BR>\n" );
document.write( "r = the common ratio.<BR>\n" );
document.write( "n = the number of terms in the sequence.<BR>\n" );
document.write( "S = a * (1-r^n) / (1-r)<BR>\n" );
document.write( "S = the sum of the terms in the sequence.<BR>\n" );
document.write( "your sequence is as follows:<BR>\n" );
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document.write( "term       number\r\n" );
document.write( "1            -9/2\r\n" );
document.write( "2             3/2\r\n" );
document.write( "3             -1/2\r\n" );
document.write( "4             1/6\r\n" );
document.write( "L             1/39366\r\n" );
document.write( "</PRE><BR>\n" );
document.write( "The last term in your sequence is equal to 1/39366<BR>\n" );
document.write( "your common ratio should be (3/2) / (-9/2)<BR>\n" );
document.write( "this is equivalent to:<BR>\n" );
document.write( "(3/2) * (2/-9) which is equivalent to:<BR>\n" );
document.write( "(3*2) / (2*(-9)) which is equivalent to:<BR>\n" );
document.write( "6 / (-18) which is equivalent to:<BR>\n" );
document.write( "-(1/3)<BR>\n" );
document.write( "if you multiply -9/2 by -1/3, you get 9/6 which is equal to 3/2<BR>\n" );
document.write( "if you multiply 3/2 by -1/3, you get -3/6 which is equal to -1/2<BR>\n" );
document.write( "if you multiply -1/2 by -1/3, you get 1/6<BR>\n" );
document.write( "the common ratio is -1/3 and it is good.<BR>\n" );
document.write( "the last term in your sequence is 1/39366<BR>\n" );
document.write( "in order to find the sum, you need to know the value of n, unless you are talking about the sum of an infinite series which is not what i understood.<BR>\n" );
document.write( "the formula for the last term in the sequence that you showed is:<BR>\n" );
document.write( "L = a*r^(n-1)<BR>\n" );
document.write( "if you try to solve this by logs, you'll run into difficulty because you can't take the log of a negative number.<BR>\n" );
document.write( "in order to use the sum formula,  you need to find the value of n and logs aren't any good.<BR>\n" );
document.write( "we'll do this by iteration to see if we can come up with the value of n.<BR>\n" );
document.write( "let's see how this will work.<BR>\n" );
document.write( "we know that the last term in our sequence is equal to 1/39366<BR>\n" );
document.write( "we work our way up in the sequence until we get that number and then we'll have the value of n.<BR>\n" );
document.write( "we start with the first value of (-9/2)<BR>\n" );
document.write( "the common ratio is (-1/3)<BR>\n" );
document.write( "formula we'll be using is:<BR>\n" );
document.write( "L = (-9/2) * (-1/3) ^ (n-1)<BR>\n" );
document.write( "we work our way up as follows:<BR>\n" );
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document.write( "n-1       L \r\n" );
document.write( "0         = (-9/2)\r\n" );
document.write( "1         = (-9/2) * (-1/3) = (-9*-1)/(2*3) = 9/6 = 3/2\r\n" );
document.write( "...\r\n" );
document.write( "10        = (-9/2) * (-1/3)^10 = (-9/2) * (-1/59049) = (-9*-1)/(2*59049)\r\n" );
document.write( "          = 9/118098 = 1/13122\r\n" );
document.write( "11        = (-9/2) * (-1/3)^11 = (-9/2) * (-1/177147) = (-9*-1) / (2*177147)\r\n" );
document.write( "          = 9/354294 = 1/39366 *****\r\n" );
document.write( "</PRE><BR>\n" );
document.write( "we found the value of n-1 and it is equal to 11.<BR>\n" );
document.write( "this means that the value of n is equal to 12.<BR>\n" );
document.write( "we have:<BR>\n" );
document.write( "n = 12<BR>\n" );
document.write( "now we can plug the values into the sum formula to get the sum of the sequence.<BR>\n" );
document.write( "the sum of this geometric series is given by the formula:<BR>\n" );
document.write( "S = a*(1-r^n)/(1-r)<BR>\n" );
document.write( "replacing with known values, we get:<BR>\n" );
document.write( "S = (-9/2) * (1 - (-1/3)^12) / (1-(-1/3)<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "S = (-9/2) * ( 1 - (1/531441) / (4/3) which becomes:<BR>\n" );
document.write( "S = (-9/2) * (531440/531441) / (4/3) which becomes:<BR>\n" );
document.write( "S = (-9/2) * (531440/531441) * (3/4) which becomes:<BR>\n" );
document.write( "S = (-9 * 531440 * 3) / (2 * 531441 * 4) which becomes:<BR>\n" );
document.write( "S = -3.374993649<BR>\n" );
document.write( "Because this is so messy, I confirmed the results using Excel as shown below:<BR>\n" );
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document.write( "a =             -4.5\r\n" );
document.write( "r = 	        -0.333333333\r\n" );
document.write( "formula =	a*r^(n-1)\r\n" );
document.write( "	\r\n" );
document.write( "n	        results of formula for each n\r\n" );
document.write( "1	        -4.5\r\n" );
document.write( "2	        1.5\r\n" );
document.write( "3	       -0.5\r\n" );
document.write( "4	        0.166666667\r\n" );
document.write( "5	       -0.055555556\r\n" );
document.write( "6	        0.018518519\r\n" );
document.write( "7	       -0.00617284\r\n" );
document.write( "8	        0.002057613\r\n" );
document.write( "9	       -0.000685871\r\n" );
document.write( "10	        0.000228624\r\n" );
document.write( "11	       -7.62079E-05\r\n" );
document.write( "12 	        2.54026E-05\r\n" );
document.write( "	        sum of all results\r\n" );
document.write( "total	       -3.374993649\r\n" );
document.write( "</PRE><BR>\n" );
document.write( "the results are confirmed so the answer looks good.<BR>\n" );
document.write( "you caught a bear this time.<BR>\n" );
document.write( "not only did they give you off the wall numbers to work with that are exceptionally difficult to do by hand, but they also hit you with a zinger in that you couldn't even use logs to find the value of n because you were dealing with logs of negative numbers which is not allowed.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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