document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=29589>Question 29589</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am having trouble with this problem, could you please help me.\r<BR>\n" );
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document.write( "3)	Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27&#8230; to find the following.\r<BR>\n" );
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document.write( "PROBLEMS:<BR>\n" );
document.write( "b)	Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms?\r<BR>\n" );
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document.write( "c)	Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms?\r<BR>\n" );
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document.write( "d)	What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?\r<BR>\n" );
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document.write( "Thank you in advance for all your help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #16597 by </B><A HREF=/tutors/aboutme.mpl?userid=sdmmadam@yahoo.com><B>sdmmadam@yahoo.com(530)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=sdmmadam@yahoo.com><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=16597\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27&#8230; to find the following.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "PROBLEMS:<BR>\n" );
document.write( "b)	Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms?<BR>\n" );
document.write( "c)	Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms?<BR>\n" );
document.write( "d)	What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?\r<BR>\n" );
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document.write( "The general geometric progression is  <BR>\n" );
document.write( "a, ar,ar^2,ar^3,ar^4, .....,ar^(n-1),......<BR>\n" );
document.write( "The given geometric progression is   1, 1/3, 1/9 , 1/27&#8230;....<BR>\n" );
document.write( "So here   a = 1 and r = (1/3)<BR>\n" );
document.write( "The formula for the sum to n terms is <BR>\n" );
document.write( "S = a[1-(r^n)]/(1-r)     ----(I)<BR>\n" );
document.write( "1)To find the sum of the first 10 terms of    1, 1/3, 1/9 , 1/27&#8230;....<BR>\n" );
document.write( "Putting a = 1, r= (1/3) and n =10<BR>\n" );
document.write( "S= a/(1-r)[1-(r^n)]<BR>\n" );
document.write( "=1/[1-(1/3)]X[1-(1/3)10]<BR>\n" );
document.write( "=[1/(2/3)[1-1/(3^10)]   (using (u/v)^p = (u^p)/(v^p) and here  u = 1and v = 3 and p= 10 )<BR>\n" );
document.write( "=(3/2)[3^(10) - 1]/(3^10)<BR>\n" );
document.write( "= [3^(10) - 1]/[2X(3^9)]  (cancelling 3 in the nr and in the dr)<BR>\n" );
document.write( "2)To find the sum of the first 12 terms of    1, 1/3, 1/9 , 1/27&#8230;....<BR>\n" );
document.write( "Putting a = 1, r= (1/3) and n =12<BR>\n" );
document.write( "S= a/(1-r)[1-(r^n)]<BR>\n" );
document.write( "=1/[1-(1/3)]X[1-(1/3)12]<BR>\n" );
document.write( "=[1/(2/3)[1-1/(3^12)]  (using (u/v)^p = (u^p)/(v^p) and here  u = 1and v = 3 and p= 12 )<BR>\n" );
document.write( "=(3/2)[3^(12) - 1]/(3^12)<BR>\n" );
document.write( "= [3^(12) - 1]/[2X(3^11)]  (cancelling 3 in the nr and in the dr)<BR>\n" );
document.write( "d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?<BR>\n" );
document.write( "If  the number n keeps increasing and  for r, the common ratio < 1 <BR>\n" );
document.write( "as in this example r = (1/3) which is less than 1<BR>\n" );
document.write( "r^n keeps decreasing  until fianlly when n is very large,so large that it tends to infinity,<BR>\n" );
document.write( "then (r^n) tends to zero which implies [1-(r^n] tends to (1-0) = 1  ----(*)<BR>\n" );
document.write( "And the sum to infinity of this above series 1+1/3+1/9 +1/27&#8230;...<BR>\n" );
document.write( "S = a[1-(r^n)]/(1-r)     ----(I)<BR>\n" );
document.write( "becomes  S = aX(1)/(1-r)<BR>\n" );
document.write( " = 1/(1-r) <BR>\n" );
document.write( "= 1/[1-(1/3)]<BR>\n" );
document.write( "= 1/(2/3)<BR>\n" );
document.write( "=3/2\r<BR>\n" );
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document.write( "Note: I don't follow exactly this statement:what number does it appear that the sum will always be smaller than?<BR>\n" );
document.write( " </SPAN>\n" );
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