document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1203303>Question 1203303</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I was wondering if I could get some help on this problem?<BR>\n" );
document.write( "If f(n) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=log%28n%29%2Flog%282006n-n%5E2%29 ALIGN=MIDDLE  ALT=\"log%28n%29%2Flog%282006n-n%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, find f(1) + f(2) + f(3)+...+ f(2005)<BR>\n" );
document.write( "All I can figure out is the split the logarithm on the bottom into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=log%28n%29+%2B+log%282006+-+n%29 ALIGN=MIDDLE  ALT=\"log%28n%29+%2B+log%282006+-+n%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #838752 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=838752\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Claim:<BR>\n" );
document.write( "f(n) + f(2006-n) = 1 where n is an integer in the set {1,2,3,...,2005}\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Here's a proof:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28n%29+=+log%28%28n%29%29%2Flog%28%282006n-n%5E2%29%29 ALIGN=MIDDLE  ALT=\"f%28n%29+=+log%28%28n%29%29%2Flog%28%282006n-n%5E2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28n%29+=+log%28%28n%29%29%2Flog%28%28n%282006-n%29%29%29 ALIGN=MIDDLE  ALT=\"f%28n%29+=+log%28%28n%29%29%2Flog%28%28n%282006-n%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28%282006-n%29%282006-%282006-n%29%29%29%29 ALIGN=MIDDLE  ALT=\"f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28%282006-n%29%282006-%282006-n%29%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28%282006-n%29%282006-2006%2Bn%29%29%29 ALIGN=MIDDLE  ALT=\"f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28%282006-n%29%282006-2006%2Bn%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28%282006-n%29n%29%29 ALIGN=MIDDLE  ALT=\"f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28%282006-n%29n%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28n%282006-n%29%29%29 ALIGN=MIDDLE  ALT=\"f%282006-n%29+=+log%28%282006-n%29%29%2Flog%28%28n%282006-n%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Both f(n) and f(2006-n) have the same denominator.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "It allows us to add the numerators and use the log rule log(A)+log(B) = log(AB)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28n%29%2Bf%282006-n%29+=+log%28%28n%29%29%2Flog%28%28n%282006-n%29%29%29%2Blog%28%282006-n%29%29%2Flog%28%28n%282006-n%29%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28n%29%2Bf%282006-n%29+=+%28log%28%28n%29%29%2Blog%28%282006-n%29%29%29%2Flog%28%28n%282006-n%29%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28n%29%2Bf%282006-n%29+=+%28log%28%28n%282006-n%29%29%29%29%2Flog%28%28n%282006-n%29%29%29 ALIGN=MIDDLE  ALT=\"f%28n%29%2Bf%282006-n%29+=+%28log%28%28n%282006-n%29%29%29%29%2Flog%28%28n%282006-n%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28n%29%2Bf%282006-n%29+=+1 ALIGN=MIDDLE  ALT=\"f%28n%29%2Bf%282006-n%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Then using that formula we can write<BR>\n" );
document.write( "f(1) + f(2005) = 1<BR>\n" );
document.write( "f(2) + f(2004) = 1<BR>\n" );
document.write( "f(3) + f(2003) = 1<BR>\n" );
document.write( "f(4) + f(2002) = 1<BR>\n" );
document.write( "...<BR>\n" );
document.write( "...<BR>\n" );
document.write( "f(2002) + f(4) = 1<BR>\n" );
document.write( "f(2003) + f(3) = 1<BR>\n" );
document.write( "f(2004) + f(2) = 1<BR>\n" );
document.write( "f(2005) + f(1) = 1<BR>\n" );
document.write( "We have 2005 pairs of values. Each pair adds to 1.<BR>\n" );
document.write( "We have a sum of 1+1+1+...+1 = 2005*1 = 2005\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "This appears it could be the final answer. However, it is not. Pay close attention that double-counting has taken place here.<BR>\n" );
document.write( "For instance, f(1)+f(2005) would be treated the same as f(2005)+f(1).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "To correct this error, divide by 2.<BR>\n" );
document.write( "2005/2 = <font color=red>1002.5</font>\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Put another way, if we add straight down through all 2005 equations and do a bit of algebraic housekeeping, the left hand sides combine to:<BR>\n" );
document.write( "2*( f(1)+f(2)+f(3)+...+f(2005) )<BR>\n" );
document.write( "while the right hand sides combine to 1+1+1+..+1, aka 2005 copies of '1' being added.<BR>\n" );
document.write( "Therefore, we have the equation<BR>\n" );
document.write( "2*( f(1)+f(2)+f(3)+...+f(2005) ) = 2005<BR>\n" );
document.write( "which rearranges to<BR>\n" );
document.write( "f(1)+f(2)+f(3)+...+f(2005) = 2005/2 = <font color=red>1002.5</font>\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<font color=black size=4>Answer:</font><BR>\n" );
document.write( "f(1) + f(2) + f(3)+...+ f(2005) = <font color=red size=4>1002.5</font><BR>\n" );
document.write( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );