document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209812>Question 1209812</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Simplify \frac{1 + 3 + 5 + ... + 1999 + 2001 + 2003}{2 + 4 + 6 + ... + 2000 + 2002 + 2004 + 2006 + 2008 + 2010}. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850652 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850652\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $N = 1 + 3 + 5 + \dots + 2003$ and $D = 2 + 4 + 6 + \dots + 2010$.<BR>\n" );
document.write( "We want to find $\frac{N}{D}$.\r<BR>\n" );
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document.write( "The numerator $N$ is the sum of an arithmetic progression with first term $a_1 = 1$, last term $a_n = 2003$, and common difference $d = 2$.<BR>\n" );
document.write( "To find the number of terms $n$, we use the formula $a_n = a_1 + (n-1)d$:<BR>\n" );
document.write( "$2003 = 1 + (n-1)2$<BR>\n" );
document.write( "$2002 = 2(n-1)$<BR>\n" );
document.write( "$1001 = n-1$<BR>\n" );
document.write( "$n = 1002$<BR>\n" );
document.write( "The sum of an arithmetic series is given by $S_n = \frac{n(a_1 + a_n)}{2}$.<BR>\n" );
document.write( "So, $N = \frac{1002(1+2003)}{2} = \frac{1002(2004)}{2} = 1002(1002) = 1004004$.\r<BR>\n" );
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document.write( "The denominator $D$ is the sum of an arithmetic progression with first term $b_1 = 2$, last term $b_m = 2010$, and common difference $d = 2$.<BR>\n" );
document.write( "To find the number of terms $m$, we use the formula $b_m = b_1 + (m-1)d$:<BR>\n" );
document.write( "$2010 = 2 + (m-1)2$<BR>\n" );
document.write( "$2008 = 2(m-1)$<BR>\n" );
document.write( "$1004 = m-1$<BR>\n" );
document.write( "$m = 1005$<BR>\n" );
document.write( "The sum of an arithmetic series is given by $S_m = \frac{m(b_1 + b_m)}{2}$.<BR>\n" );
document.write( "So, $D = \frac{1005(2+2010)}{2} = \frac{1005(2012)}{2} = 1005(1006) = 1011030$.\r<BR>\n" );
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document.write( "Then $\frac{N}{D} = \frac{1004004}{1011030}$.<BR>\n" );
document.write( "We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor.<BR>\n" );
document.write( "Let's divide both by 6:<BR>\n" );
document.write( "$\frac{1004004}{6} = 167334$<BR>\n" );
document.write( "$\frac{1011030}{6} = 168505$<BR>\n" );
document.write( "So $\frac{N}{D} = \frac{167334}{168505}$.<BR>\n" );
document.write( "Let's check if there are any more common factors.<BR>\n" );
document.write( "We can use the Euclidean algorithm to find the gcd of 167334 and 168505.<BR>\n" );
document.write( "$168505 = 1(167334) + 1171$<BR>\n" );
document.write( "$167334 = 142(1171) + 832$<BR>\n" );
document.write( "$1171 = 1(832) + 339$<BR>\n" );
document.write( "$832 = 2(339) + 154$<BR>\n" );
document.write( "$339 = 2(154) + 31$<BR>\n" );
document.write( "$154 = 4(31) + 30$<BR>\n" );
document.write( "$31 = 1(30) + 1$<BR>\n" );
document.write( "The gcd is 1, so the fraction is already in its simplest form.\r<BR>\n" );
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document.write( "Thus, $\frac{N}{D} = \frac{1004004}{1011030} = \frac{167334}{168505}$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{\frac{167334}{168505}}$<BR>\n" );
document.write( " </SPAN>\n" );
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