document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1061781>Question 1061781</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Sam deposits 100 dollars in his account on Tuesday. If he doubles the amount he deposits every day, what is the equation that will determine how much is in his account? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #676534 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=676534\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Second day: 200<BR>\n" );
document.write( "first day: 100 <BR>\n" );
document.write( "So we have a series with a ratio of 200/100 = 2<BR>\n" );
document.write( "Now, the sum of a geometric series S with n terms and with a ratio r is:<BR>\n" );
document.write( "S = (first term)(1-r^n)/(1-r)<BR>\n" );
document.write( "Since we have a first term 100 and a ratio of 2, to find the sum after n days we do this:<BR>\n" );
document.write( "= 100(1-2^n)/(1-2)<BR>\n" );
document.write( "-.-.-.-.-.-.-.-.<BR>\n" );
document.write( "Let's test this formula. Let's say that he's been saving for 4 days:<BR>\n" );
document.write( "100(1-2^4)/(1-2) = 1,500<BR>\n" );
document.write( "Compare with this:<BR>\n" );
document.write( "Day 1 = 200-100 = 100<BR>\n" );
document.write( "Day 2 = 400-100 = 300<BR>\n" );
document.write( "Day 3 = 800-100 = 700<BR>\n" );
document.write( "Day 4 = 1600-100 = 1500<BR>\n" );
document.write( "So our formula works. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );